4. Computing Limits
Exercises
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\(\displaystyle\lim_{t\to23}30\)
What is the limit of a constant?
\(\displaystyle\lim_{t\to23}30=30\)
\(30\) is a constant. So: \[ \lim_{t\to23}30 =30 \]
tj
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\(\displaystyle\lim_{x\to2}5^x\)
Use the Exponential Law.
\(\displaystyle\lim_{x\to2}5^x=25\)
We use the Exponential Law: \[\begin{aligned} \lim_{x\to2}5^x &=5^{\lim\limits_{x\to2}x} \\ &=5^2 =25 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to-3}(x+x^2)\)
Use the addition and power laws.
\(\displaystyle\lim_{x\to-3}(x+x^2)=6\)
We use the addition law: \[\begin{aligned} \lim_{x\to-3}(x+x^2) &=\lim_{x\to-3}x+\lim_{x\to-3}x^2 \\ &=-3+\lim_{x\to-3}x^2 \end{aligned}\] Now, we use the power law: \[\begin{aligned} \lim_{x\to-3}(x+x^2) &=-3+(\lim_{x\to-3}x)^2 \\ &=-3+(-3)^2 \\ &=-3+9 =6 \end{aligned}\]
tj
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\(\displaystyle\lim_{t\to6}\dfrac{t^2+1}{t}\)
Use the quotient law, the addition law, and the power law.
\(\displaystyle\lim_{t\to6}\dfrac{t^2+1}{t}=\dfrac{37}{6}\)
We use the quotient law: \[\begin{aligned} \lim_{t\to6}\dfrac{t^2+1}{t} &=\dfrac{\lim\limits_{t\to6}(t^2+1)}{\lim\limits_{t\to6}t} \\ &= \dfrac{\lim\limits_{t\to6}(t^2+1)}{6} \end{aligned}\] Notice the quotient law is valid because the denominator is \(\lim\limits_{t\to6}t=6\ne0\). Now, we use the addition law: \[\begin{aligned} \lim_{t\to6}\dfrac{t^2+1}{t} &=\dfrac{\lim\limits_{t\to6}t^2+\lim\limits_{t\to6}1}{6} \\ &=\dfrac{\lim\limits_{t\to6}t^2+1}{6} \end{aligned}\] Finally, we use the power law: \[\begin{aligned} \lim_{t\to6}\dfrac{t^2+1}{t} &=\dfrac{(\lim\limits_{t\to6}t)^2+1}{6} \\ &=\dfrac{6^2+1}{6} =\dfrac{37}{6} \end{aligned}\]
tj
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\(\displaystyle\lim_{y\to0}\dfrac{1}{y^2+1}\)
Use the quotient rule and the power rule.
\(\displaystyle\lim_{y\to0}\dfrac{1}{y^2+1}=1\)
We use the quotient rule: \[ \lim_{y\to0}\dfrac{1}{y^2+1} =\dfrac{1}{\lim\limits_{y\to0}(y^2+1)} \] Now, we use the power rule: \[\begin{aligned} \lim_{y\to0}\dfrac{1}{y^2+1} &=\dfrac{1}{\lim\limits_{y\to0}(y^2)+1}\\ &=\dfrac{1}{0^2+1} =\dfrac{1}{1} =1 \end{aligned}\] Notice the quotient law was valid because the denominator came out to be \(\displaystyle \lim_{y\to0}(y^2+1)=1\ne0\).
tj
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\(\displaystyle\lim_{y\to4}3y\)
Use the constant multiple rule.
\(\displaystyle\lim_{y\to4}3y=12\)
We use the constant multiple rule: \[\begin{aligned} \lim_{y\to4}3y =3\cdot\lim_{y\to4}y =3\cdot4 =12 \end{aligned}\]
tj
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If \(\displaystyle\lim_{x\to3}f(x)=7\) and \(\displaystyle \lim_{x\to3}g(x)=4\), compute: \[\lim_{x\to3} \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2} \]
Use the Limit Laws.
\(\displaystyle\lim_{x\to3} \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}=\dfrac{17}{15}\)
We use the quotient rule: \[\begin{aligned} \lim_{x\to3}& \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}\\ &=\dfrac{\lim\limits_{x\to3}\left(g(x)^2-4f(x)g(x)+3f(x)^2\right)} {\lim\limits_{x\to3}\left(-2g(x)^2+f(x)g(x)+f(x)^2\right)} \end{aligned}\] We use the addition and subtraction rules and the constant multiple rule: \[\begin{aligned} \lim_{x\to3}& \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}\\ &=\dfrac{\lim\limits_{x\to3}\left(g(x)^2\right)-4\lim\limits_{x\to3}\left( f(x)g(x)\right)+3\lim\limits_{x\to3}\left(f(x)^2\right)} {-2\lim\limits_{x\to3}\left(g(x)^2\right)+ \lim\limits_{x\to3}\left(f(x)g(x)\right)+\lim\limits_{x\to3}\left( f(x)^2\right)} \end{aligned}\] We use the power and product laws: \[\begin{aligned} \lim_{x\to3}& \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}\\ &=\dfrac{16-112+147}{-32+28+49}=\dfrac{51}{45}=\dfrac{17}{15} \end{aligned}\] Since the limit of the denominator came out to be \(45\) which is non-zero, the quotient rule used at the start is valid.
tj
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\(\displaystyle \lim_{x\to2}\dfrac{x^2+2x-8}{x-2}\)
Factor the numerator and cancel.
\(\displaystyle \lim_{x\to2}\dfrac{x^2+2x-8}{x-2}=6\)
We factor the numerator and cancel: \[\begin{aligned} \lim_{x\to2}\dfrac{x^2+2x-8}{x-2} &=\lim_{x\to2}\dfrac{(x-2)(x+4)}{x-2} \\ &=\lim_{x\to2}(x+4)=6 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to-2}\dfrac{x^2-x-6}{x+2}\)
Factor the numerator and cancel.
\(\displaystyle \lim_{x\to-2}\dfrac{x^2-x-6}{x+2}=-5\)
We factor the numerator and cancel: \[\begin{aligned} \lim_{x\to-2}\dfrac{x^2-x-6}{x+2} &=\lim_{x\to-2}\dfrac{(x+2)(x-3)}{x+2} \\ &=\lim_{x\to-2}(x-3)=-5 \end{aligned}\]
tj
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\(\displaystyle\lim_{k\to7}\dfrac{k^2-4k-21}{k-7}\)
Factor the numerator and cancel.
\(\displaystyle\lim_{k\to7}\dfrac{k^2-4k-21}{k-7}=10\)
We expand the numerator and cancel: \[\begin{aligned} \lim_{k\to7}\dfrac{k^2-4k-21}{k-7} &=\lim_{k\to7}\dfrac{(k+3)(k-7)}{k-7}\\ &=\lim_{k\to7}(k+3) =10 \end{aligned}\]
tj
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\(\displaystyle\lim_{k\to-5}\dfrac{k+5}{k^2+2k-15}\)
Factor the denominator and cancel.
\(\displaystyle\lim_{k\to-5}\dfrac{k+5}{k^2+2k-15} =-\dfrac{1}{8}\)
We factor the denominator and cancel: \[\begin{aligned} \lim_{k\to-5}\dfrac{k+5}{k^2+2k-15} &=\lim_{k\to-5}\dfrac{k+5}{(k+5)(k-3)}\\ &=\lim_{k\to-5}\dfrac{1}{k-3} =-\dfrac{1}{8} \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to21}\dfrac{x^2-441}{x-21}\)
Factor the numerator and cancel.
\(\lim_{x\to21}\dfrac{x^2-441}{x-21}=42\)
We factor the numerator and cancel: \[\begin{aligned} \lim_{x\to21}&\dfrac{x^2-441}{x-21} =\lim_{x\to21}\dfrac{(x+21)(x-21)}{x-21}\\ &=\lim_{x\to21}(x+21) =21+21 =42 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0}\dfrac{(x-4)^2-16}{x}\)
Expand the numerator and cancel.
\(\displaystyle \lim_{x\to0}\dfrac{(x-4)^2-16}{x} =-8\)
We expand the numerator and cancel: \[\begin{aligned} \lim_{x\to0}&\dfrac{(x-4)^2-16}{x} =\lim_{x\to0}\dfrac{x^2-8x}{x} \\ &=\lim_{x\to0}(x-8) =-8 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to3}\dfrac{(5-x)^2-4}{x-3}\)
Expand and then factor the numerator and cancel.
\(\displaystyle \lim_{x\to3}\dfrac{(5-x)^2-4}{x-3}=-4\)
We expand and then factor the numerator and cancel: \[\begin{aligned} \lim_{x\to3}&\dfrac{(5-x)^2-4}{x-3} =\lim_{x\to3}\dfrac{x^2-10x+21}{x-3} \\ &=\lim_{x\to3}\dfrac{(x-3)(x-7)}{x-3} =\lim_{x\to3}(x-7) =-4 \end{aligned}\]
tj
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\(\displaystyle \lim_{h\to0}\dfrac{(4+h)^2-16}{h}\)
Expand the numerator and cancel.
\(\displaystyle \lim_{h\to0}\dfrac{(4+h)^2-16}{h}=8\)
We expand the numerator and cancel: \[\begin{aligned} \lim_{h\to0}&\dfrac{(4+h)^2-16}{h} =\lim_{h\to0}\dfrac{8h+h^2}{h} \\ &=\lim_{x\to0}(8+h) =8 \end{aligned}\]
tj
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\(\displaystyle \lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}\)
Expand the numerator and cancel.
\(\displaystyle \lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}=2x\)
We expand the numerator and cancel: \[\begin{aligned} \lim_{h\to0}&\dfrac{(x+h)^2-x^2}{h} =\lim_{h\to0}\dfrac{2hx+h^2}{h} \\ &=\lim_{h\to0}(2x+h) =2x \end{aligned}\]
tj
We will shortly learn that this is one way to compute the derivative of \(x^2\).
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\(\displaystyle \lim_{x\to5} \dfrac{ (x-10)^2 - 25 }{ x - 5 } \)
Expand and then factor the numerator and cancel.
\(\displaystyle\lim_{x\to5} \dfrac{ (x-10)^2 - 25 }{ x - 5 } = -10 \)
We expand and then factor the numerator and cancel: \[\begin{aligned} \lim_{x\to5}&\dfrac{ (x-10)^2 - 25 }{ x - 5 } =\lim_{x\to5} \dfrac{ x^2 - 20x + 100 - 25 }{ x - 5 }\\ &=\lim_{x\to5} \dfrac{ x^2 - 20x + 75 }{ x - 5 } =\lim_{x\to5} \dfrac{ (x - 5)(x - 15) }{ x - 5 }\\ &=\lim_{x\to5} (x-15) =5-15 =-10 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0} \left(\dfrac{8x^3}{3}-4x^2\right)\left(\dfrac{2}{x^2}+6\right) \)
Try multiplying it out.
\( \lim_{x\to0} \left(\dfrac{8x^3}{3}-4x^2\right)\left(\dfrac{2}{x^2}+6\right) =-8\)
We can solve this problem by multiplying the terms out. \[\begin{aligned} \lim_{x\to0} \left(\dfrac{8x^3}{3}-4x^2\right)\left(\dfrac{2}{x^2}+6\right) &= \lim_{x\to0} \left(\dfrac{16x}{3}+16x^3-8-24x^2\right) \\ &= \left(0+0-8-0\right) = -8 \end{aligned}\]
by
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\(\displaystyle\lim_{x\to0}\dfrac{8x^2-3x}{x^2+5x}\)
Divide both the numerator and the denominator by the largest term in the denominator. For \(x\) close to \(0\), which is larger: \(x\) or \(x^2\)?
\(\displaystyle\lim_{x\to0}\dfrac{8x^2-3x}{x^2+5x}=-\dfrac{3}{5}\)
For \(x\) close to \(0\), which is larger: \(x\) or \(x^2\)? At \(x=.1\), we have \(x^2=.01\) and so \(x\) is larger. We divide both the numerator and the denominator by the largest term in the denominator which is \(x\): \[\begin{aligned} \lim_{x\to0}&\dfrac{8x^2-3x}{x^2+5x} =\lim_{x\to0}\dfrac{(8x^2-3x)}{(x^2+5x)} \dfrac{\dfrac{1}{x}}{\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{8x-3}{x+5} =-\,\dfrac{3}{5} \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}\dfrac{x^2+3x}{2x^2-7x}\)
Divide both the numerator and the denominator by the largest term in the denominator. For \(x\) colse to \(0\), which is larger: \(x\) or \(x^2\)?
\(\displaystyle\lim_{x\to0}\dfrac{x^2+3x}{2x^2-7x}=-\,\dfrac{3}{7}\)
For \(x\) close to \(0\), which is larger: \(x\) or \(x^2\)? At \(x=.1\), we have \(x^2=.01\) and so \(x\) is larger. We divide both the numerator and the denominator by the largest term in the denominator: \[\begin{aligned} \lim_{x\to0}\dfrac{x^2+3x}{2x^2-7x} &=\lim_{x\to0}\dfrac{(x^2+3x)\cdot\dfrac{1}{x}}{(2x^2-7x)\cdot\dfrac{1}{x} }\\ &=\lim_{x\to0}\dfrac{x+3}{2x-7} =-\,\dfrac{3}{7} \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}\dfrac{6x-3x^4}{3x+9x^3}\)
Divide the numerator and denominator by largest term in the denominator.
\(\displaystyle\lim_{x\to0}\dfrac{6x-3x^4}{3x+9x^3}=2\)
We divide the numerator and denominator by largest term in the denominator: \[\begin{aligned} \lim_{x\to0}&\dfrac{6x-3x^4}{3x+9x^3} =\lim_{x\to0}\dfrac{(6x-3x^4)\cdot\dfrac{1}{x}}{(3x+9x^3)\cdot\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{6-3x^3}{3+9x^2} =\dfrac{6}{3} =2 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}\dfrac{4x^2-3x^5}{3x+8x^2}\)
Divide the numerator and denominator by largest term in the denominator.
\(\displaystyle\lim_{x\to0}\dfrac{4x^2-3x^5}{3x+8x^2}=0\)
We divide the numerator and denominator by largest term in the denominator: \[\begin{aligned} \lim_{x\to0}\dfrac{4x^2-3x^5}{3x+8x^2} &=\lim_{x\to0}\dfrac{(4x^2-3x^5)\cdot\dfrac{1}{x}}{(3x+8x^2)\cdot\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{4x-3x^4}{3+8x} =\dfrac{0}{3} =0 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0}\dfrac{3x^3-7x^4}{4x^3+x^4} \)
Divide the numerator and denominator by largest term in the denominator. For \(x\) colse to \(0\), which is larger: \(x^3\) or \(x^4\)?
\(\displaystyle \lim_{x\to0}\dfrac{3x^3-7x^4}{4x^3+x^4} = \dfrac{3}{4} \)
For \(x\) close to \(0\), which is larger: \(x^3\) or \(x^4\)? At \(x=.1\), we have \(x^3=.001\) and \(x^4=.0001\). So \(x^3\) is larger. We divide the numerator and denominator by largest term in the denominator: \[\begin{aligned} \lim_{x\to0}\dfrac{3x^3-7x^4}{4x^3+x^4} &=\lim_{x\to0}\dfrac{(3x^3-7x^4) \cdot \dfrac{1}{x^3} }{ ( 4x^3+x^4 ) \cdot \dfrac{1}{x^3}}\\ &=\lim_{x\to0}\dfrac{3-7x}{4+x} =\dfrac{3}{4} \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0} \left( \dfrac{1}{x}-\dfrac{1}{x+x^2+x^3} \right) \)
Put the terms over a common denominator.
\(\displaystyle \lim_{x\to0} \left( \dfrac{1}{x}-\dfrac{1}{x+x^2+x^3} \right) = 1 \)
We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to0} &\left( \dfrac{1}{x}-\dfrac{1}{x+x^2+x^3} \right) \\ &=\lim_{x\to0} \left(\dfrac{1+x+x^2}{x+x^2+x^3}-\dfrac{1}{x+x^2+x^3}\right)\\ &=\lim_{x\to0}\dfrac{x+x^2}{x+x^2+x^3} =\lim_{x\to0}\dfrac{1+x}{1+x+x^2} =1 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0} \left( \dfrac{1}{3x+x^2}-\dfrac{1}{3x} \right) \)
Put the terms over a common denominator.
\(\displaystyle \lim_{x\to0} \left( \dfrac{1}{3x+x^2}-\dfrac{1}{3x} \right) = -\dfrac{1}{9} \)
We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to0} &\left( \dfrac{1}{3x+x^2}-\dfrac{1}{3x} \right) =\lim_{x\to0} \left( \dfrac{3}{9x+3x^2}-\dfrac{3+x}{9x+3x^2} \right)\\ &=\lim_{x\to0} \left( -\,\dfrac{x}{9x+3x^2}\right) =\lim_{x\to0} \left( -\,\dfrac{1}{9+3x} \right) =-\dfrac{1}{9} \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}\left(\dfrac{1}{x}-\dfrac{1}{x+2x^2}\right)\)
Put the terms over a common denominator.
\(\displaystyle\lim_{x\to0}\left(\dfrac{1}{x}-\dfrac{1}{x+2x^2}\right)=2\)
We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to0}&\left(\dfrac{1}{x}-\dfrac{1}{x+2x^2}\right) =\lim_{x\to0}\left(\dfrac{1+2x}{x+2x^2}-\dfrac{1}{x+2x^2}\right) \\ &=\lim_{x\to0}\dfrac{2x}{x+2x^2} =\lim_{x\to0}\dfrac{2}{1+2x} =2 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0} \left(\dfrac{3+2x^3}{x+4x^3}-\dfrac{x^3+6}{2x+x^3}\right)\)
Put the terms over a common denominator
\(\displaystyle\lim_{x\to0} \left( \dfrac{3+2x^3}{x+4x^3} - \dfrac{x^3+6}{2x+x^3} \right) = 0 \)
We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to0}& \left( \dfrac{3+2x^3}{x+4x^3} - \dfrac{x^3+6}{2x+x^3} \right) \\ &=\lim_{x\to0} \left( \dfrac{(3+2x^3)(2x+x^3)}{(x+4x^3)(2x+x^3)} - \dfrac{(x^3+6)(x+4x^3)}{(2x+x^3)(x+4x^3)} \right)\\ &=\lim_{x\to0} \left( \dfrac{6x+3x^3+4x^4+2x^6}{2x^2+9x^4+4x^6} - \dfrac{x^4+4x^6+6x+24x^3}{2x^2+9x^4+4x^6} \right)\\ &=\lim_{x\to0} \dfrac{-21x^3+3x^4-2x^6}{2x^2+9x^4+4x^6} =\lim_{x\to0} \dfrac{-21x+3x^2-2x^4}{2+9x^2+4x^4} =0 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to0} \left(\dfrac{8x^2-7x^3}{x^2+3x^3}+\dfrac{9x^2+9x^3}{4x+x^3}\right)\)
Put the terms over a common denominator or try cancelling first.
\(\displaystyle \lim_{x\to0}\left( \dfrac{8x^2-7x^3}{x^2+3x^3} + \dfrac{9x^2+9x^3}{4x+x^3} \right)=8 \)
We simplify before putting the terms over a common denominator: \[ \lim_{x\to0}\left( \dfrac{8x^2-7x^3}{x^2+3x^3} + \dfrac{9x^2+9x^3}{4x+x^3} \right) =\lim_{x\to0}\left( \dfrac{8-7x}{1+3x} + \dfrac{9x+9x^2}{4+x^2} \right) \] Turns out we don't need to put it over a common denominator. \[ \lim_{x\to0}\left( \dfrac{8x^2-7x^3}{x^2+3x^3} + \dfrac{9x^2+9x^3}{4x+x^3} \right) =\dfrac{8-7\cdot0}{1+3\cdot0} + \dfrac{9\cdot0+9\cdot0^2}{4+0^2} =8 \]
tj
It will also work if you put the terms over a common denominator but it is much harder!
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\(\displaystyle \lim_{x\to3}\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } \)
Multiply by the conjugate.
\(\displaystyle \lim_{x\to3}\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } =\dfrac{1}{\sqrt{3}}\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to3}\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } &=\lim_{x\to3}\left(\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } \cdot\dfrac{ \sqrt{3x-6} + \sqrt{x} }{ \sqrt{3x-6} + \sqrt{x} }\right)\\ &=\lim_{x\to3}\dfrac{ (3x-6) - (x) }{ (x-3)(\sqrt{3x-6} + \sqrt{x}) }\\ &=\lim_{x\to3}\dfrac{ 2x - 6 }{ (x-3)(\sqrt{3x-6} + \sqrt{x}) }\\ &=\lim_{x\to3}\dfrac{ 2 }{ \sqrt{3x-6} + \sqrt{x} }\\ &=\dfrac{2}{\sqrt{3}+\sqrt{3}} =\dfrac{1}{\sqrt{3}} \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to6}\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6}\)
Multiply by the conjugate of the numerator.
Remember \((a-b)(a+b)=a^2-b^2\).\(\displaystyle \lim_{x\to6}\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6}=\dfrac{\sqrt{6}}{12}\)
We multiply by the conjugate of the numerator and then multiply out the numerator: \[\begin{aligned} \lim_{x\to6}&\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6} \\ &=\lim_{x\to6}\left(\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6} \cdot \dfrac{ \sqrt{2x-6}+\sqrt{x} }{ \sqrt{2x-6}+\sqrt{x} } \right) \\ &=\lim_{x\to0}\dfrac{(2x-6)-(x)}{(x-6)(\sqrt{2x-6}+\sqrt{x})}\\ &=\lim_{x\to6}\dfrac{1}{\sqrt{2x-6}+\sqrt{x}} =\dfrac{1}{2\sqrt{6}} =\dfrac{\sqrt{6}}{12} \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\right)\)
Multiply by the conjugate.
\(\displaystyle \lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\right)=\dfrac{1}{2}\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\right) &=\lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\cdot\dfrac{ \sqrt{1+x}+\sqrt{7-x}}{\sqrt{1+x}+\sqrt{7-x}}\right)\\ &=\lim_{x\to3}\left( \dfrac { (1+x) - (7-x) } { (x-3)( \sqrt{1+x}+ \sqrt{7-x} ) } \right)\\ &=\lim_{x\to3}\left( \dfrac { 2x-6 } { (x-3)( \sqrt{1+x}+ \sqrt{7-x} ) } \right)\\ &=\lim_{x\to3}\left( \dfrac { 2 } { \sqrt{1+x}+ \sqrt{7-x} } \right) =\dfrac{2}{4} =\dfrac{1}{2} \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to4}\dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x} \)
\(\displaystyle \lim_{x\to4}\dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x} =\dfrac{1}{\sqrt{20}} \)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to4}&\dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x} \\ &=\lim_{x\to4} \left( \dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x}\cdot \dfrac{ \sqrt{x^2+x}+\sqrt{9x-x^2} }{ \sqrt{x^2+x}+\sqrt{9x-x^2} } \right)\\ &=\lim_{x\to4} \dfrac{(x^2+x)-(9x-x^2)}{(x^2-4x)(\sqrt{x^2+x}+\sqrt{9x-x^2})}\\ &=\lim_{x\to4} \dfrac{2x^2-8x}{(x^2-4x)(\sqrt{x^2+x}+\sqrt{9x-x^2})}\\ &=\lim_{x\to4} \dfrac{2}{\sqrt{x^2+x}+\sqrt{9x-x^2}} =\dfrac{1}{\sqrt{20}} \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to4}\dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} } \)
Multiply by the conjugate.
\(\displaystyle \lim_{x\to4}\dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} } = \sqrt{5}\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to4}&\dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} }\\ &=\lim_{x\to4} \left( \dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} } \cdot \dfrac{ \sqrt{x+1} + \sqrt{9-x} }{ \sqrt{x+1} + \sqrt{9-x} }\right)\\ &=\lim_{x\to4} \left( \dfrac{ (x-4)(\sqrt{x+1} + \sqrt{9-x}) }{ (x+1) - (9-x) } \right)\\ &=\lim_{x\to4} \left( \dfrac{ (x-4)(\sqrt{x+1} + \sqrt{9-x}) }{ 2x-8 } \right)\\ &=\lim_{x\to4} \left( \dfrac{ \sqrt{x+1} + \sqrt{9-x} }{ 2 }\right) =\dfrac{\sqrt{5}+\sqrt{5}}{2} =\sqrt{5} \end{aligned}\]
tj
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\(\displaystyle \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}\)
What is the conjugate of \(1-\cos\theta\)?
\(\displaystyle \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}=2\)
We multiply by the congugate: \[\begin{aligned} \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta} &=\lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}\dfrac{1+\cos\theta}{1+\cos\theta}\\ &=\lim_{\theta\to0}\dfrac{\sin^2\theta(1+\cos\theta)}{1-\cos^2\theta} \\ &=\lim_{x\to0}(1+\cos\theta) =2 \end{aligned}\]
tj
As an alternate we can just use a trig identity: \[\begin{aligned} \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta} &=\lim_{\theta\to0}\dfrac{1-\cos^2\theta}{1-\cos\theta}\\ &=\lim_{\theta\to0}\dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}\\ &=\lim_{x\to0}(1+\cos\theta) =2 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to4}\dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} \)
\(\displaystyle \lim_{x\to4}\dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} =4\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to4}\dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} &=\lim_{x\to4} \left( \dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} \cdot \dfrac{ \sqrt{2x-4}+\sqrt{x} }{ \sqrt{2x-4}+\sqrt{x} } \right)\\ &=\lim_{x\to4} \left( \dfrac{(x-4)(\sqrt{2x-4}+\sqrt{x})} {(2x-4)-(x)}\right)\\ &=\lim_{x\to4} \left( \dfrac{(x-4)(\sqrt{2x-4}+\sqrt{x})} {x-4}\right)\\ &=\lim_{x\to4} \left(\sqrt{2x-4}+\sqrt{x} \right) =4 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to3}\dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}}\)
Multiply by the conjugate.
\(\displaystyle \lim_{x\to3}\dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}}=\dfrac{2}{3}\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to3}&\dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}} \\ &=\lim_{x\to3} \left( \dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}}\cdot \dfrac{\sqrt{2x^2-14}+\sqrt{x^2-5}}{\sqrt{2x^2-14}+\sqrt{x^2-5}} \right) \\ &=\lim_{x\to3} \dfrac{(x-3)(\sqrt{2x^2-14}+\sqrt{x^2-5})} {(2x^2-14)-(x^2-5)} \\ &=\lim_{x\to3} \dfrac{(x-3)(\sqrt{2x^2-14}+\sqrt{x^2-5})} {x^2-9} \\ &=\lim_{x\to3} \dfrac{\sqrt{2x^2-14}+\sqrt{x^2-5}} {x+3} =\dfrac{4}{6}=\dfrac{2}{3} \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}x\sin\left(\dfrac{1}{x}\right)\)
Start with \(-1 \le \sin\left(\dfrac{1}{x}\right) \le 1\).
\(\displaystyle\lim_{x\to0}x\sin\left(\dfrac{1}{x}\right)=0\)
Since \(-1 \le \sin\left(\dfrac{1}{x}\right) \le 1\), we have:
if \(x \ge 0\) then: \( -x \le x\sin\left(\dfrac{1}{x}\right) \le x \)
and if \(x \lt 0\) then: \( -x \ge x\sin\left(\dfrac{1}{x}\right) \ge x \)
In either case, \( -|x| \le x\sin\left(\dfrac{1}{x}\right) \le |x| \)
This says \( x\sin\left(\dfrac{1}{x}\right)\) is squeezed between \(-|x|\) and \(|x|\). Since, \[ \lim_{x\to0}-|x|=\lim_{x\to0}|x|=0 \] we conclude: \[ \lim_{x\to0}x\sin\left(\dfrac{1}{x}\right) =0 \] also. -
\(\displaystyle\lim_{x\to0}x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right]\)
Start with \(-1 \le \cos\left(\dfrac{1}{x}\right) \le 1\).
\(\displaystyle\lim_{x\to0}x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right] =0\)
We build up inequalities surrounding our function: \[ -1 \le \cos\left(\dfrac{1}{x}\right) \le 1 \]\[ -1 \le -\cos\left(\dfrac{1}{x}\right) \le 1 \]\[ 0 \le 1-\cos\left(\dfrac{1}{x}\right) \le 2 \]\[ 0 \le x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right] \le 2x^2 \] Using this inequality and \[ \lim_{x\to0}0=\lim_{x\to0}2x^2=0 \] the Sandwich Theorem says: \[ \lim_{x\to0}x^2\left[1-\cos\left(\dfrac{1}{x}\right)\right]=0 \] also.
tj
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\(\displaystyle\lim_{\theta\to0}\dfrac{1-\cos^4\theta}{\theta^2}\)
Factor \(1-\cos^4\theta\).
\(\displaystyle\lim_{\theta\to0}\dfrac{1-\cos^4\theta}{\theta^2}=2\)
We manipulate the limit into a form where the limit \(\displaystyle\lim_{x\to0}\dfrac{\sin x}{x}=1\) can be used: \[\begin{aligned} \lim_{\theta\to0}\dfrac{1-\cos^4\theta}{\theta^2} &=\lim_{\theta\to0}\dfrac{ (1+\cos^2\theta) ( 1-\cos^2\theta ) }{\theta^2}\\ &=\lim_{x\to0}( 1+\cos^2\theta ) \cdot \lim_{\theta\to0} \dfrac{ \sin^2\theta }{\theta^2}\\ &=2\cdot(1)^2 =2 \end{aligned}\]
tj
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\(\displaystyle \lim_{\theta\to0}\dfrac{1-\sec\theta}{\theta^2}\)
What is the conjugate of \(1-\sec\theta\)?
\(\displaystyle \lim_{\theta\to0}\dfrac{1-\sec\theta}{\theta^2}=-\dfrac{1}{2}\)
We multiply by the conjugate: \[\begin{aligned} \lim_{\theta\to0}&\dfrac{1-\sec\theta}{\theta^2} =\lim_{\theta\to0} \left( \dfrac{1-\sec\theta}{\theta^2} \cdot \dfrac{1+\sec\theta}{1+\sec\theta} \right)\\ &=\lim_{\theta\to0}\dfrac{1-\sec^2\theta}{(\theta^2)(1+\sec\theta)} =\lim_{\theta\to0}\dfrac{-\tan^2\theta}{(\theta^2)(1+\sec\theta)}\\ &=-\lim_{\theta\to0}\dfrac{\sin^2\theta}{\theta^2} \lim_{\theta\to0}\dfrac{1}{\cos^2\theta(1+\sec\theta)}\\ &=-\left(\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}\right)^2 \lim_{\theta\to0}\dfrac{1}{\cos^2\theta+\cos\theta} =-\dfrac{1}{2} \end{aligned}\]
tj
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\(\displaystyle \lim_{\theta\to0} \sqrt{ \dfrac{ \tan^2(\theta) }{ \theta } +3 } \)
Start with using the continuous function law to move the limit inside the square root.
\(\displaystyle \lim_{\theta\to0} \sqrt{ \dfrac{ \tan^2(\theta) }{\theta}+3}=\sqrt{3}\)
We use the continuous function law: \[ \lim_{\theta\to0} \sqrt{ \dfrac{ \tan^2(\theta) }{ \theta } +3} =\sqrt{ \left(\lim_{\theta\to0}\dfrac{\tan^2(\theta) }{ \theta }\right)+3} \] Now, we need to find \(\displaystyle \lim_{\theta\to0}\dfrac{\tan^2(\theta)}{ \theta }\): \[\begin{aligned} \lim_{\theta\to0}\dfrac{\tan^2(\theta)}{ \theta } &=\lim_{\theta\to0}\dfrac{\theta\cdot\sin^2(\theta)} { \theta^2\cdot\cos^2(\theta) }\\ &=\lim_{x\to0}\left(\dfrac{\sin\theta}{\theta}\right)^2 \lim_{\theta\to0}\dfrac{\theta}{\cos^2(\theta) }\\ &=1\cdot\dfrac{0}{1} =0 \end{aligned}\] Finally, we plug in \(\displaystyle \lim_{\theta\to0}\dfrac{\tan^2(\theta)}{ \theta }=0\): \[ \lim_{\theta\to0} \sqrt{ \dfrac{ \tan^2(\theta) }{ \theta } +3} =\sqrt{ \left(\lim_{\theta\to0}\dfrac{\tan^2(\theta) }{ \theta }\right)+3} =\sqrt{3} \]
tj
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\(\displaystyle\lim_{x\to\infty}\dfrac{1}{x}\)
\(\displaystyle x = x^1\).
\(\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0\)
We use one of the power laws: \[ \lim_{x\to\infty}\dfrac{1}{x} =\lim_{x\to\infty}\dfrac{1}{x^1} =0 \]
tj
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\(\displaystyle \lim_{x\to\infty}\dfrac{x}{x+1}\)
Divide by the largest term in the denominator.
\(\displaystyle \lim_{x\to\infty}\dfrac{x}{x+1}=1\)
If we plug in \(x=\infty\), we get the indeterminate form \(\dfrac{\infty} {\infty}\). This means we need to use a trick to find the limit. In this case, dividing by the largest term in the denominator works: \[ \lim_{x\to\infty}\dfrac{x}{x+1} =\lim_{x\to\infty}\dfrac{x\cdot\dfrac{1}{x}}{(x+1)\cdot\dfrac{1}{x}} =\lim_{x\to\infty}\dfrac{1}{1+\dfrac{1}{x}} =1 \]
tj
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\(\displaystyle \lim_{x \to \infty} \left(\dfrac{x^2}{2x+1}-\dfrac{x^2}{2x-2}\right) \)
\( \dfrac{-3}{4} \)
Plugging in \( x=\infty \) gives us an indeterminate form of \(\infty-\infty\) We can solve this problem by combining the fractions using the common denominator. \[\begin{aligned} \lim_{x \to \infty} \left(\dfrac{x^2}{2x+1}-\dfrac{x^2}{2x-2}\right) &= \lim_{x\to\infty} \dfrac{2x^3-2x^2-2x^3-x^2}{(2x+1)(2x-2)} \\ &= \lim_{x\to\infty} \dfrac{-3x^2}{4x^2-2x-2} \\ &=-\dfrac{3}{4} \end{aligned}\]
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\(\displaystyle \lim_{x\to\infty}\left(x-\dfrac{x^2+3}{x+4}\right) \)
Put the terms over a common denominator.
\(\displaystyle \lim_{x\to\infty}\left(x-\dfrac{x^2+3}{x+4}\right)=4\)
We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to\infty}\left(x-\dfrac{x^2+3}{x+4}\right) &=\lim_{x\to\infty}\left(\dfrac{x^2+4x}{x+4}-\dfrac{x^2+3}{x+4}\right)\\ &=\lim_{x\to\infty}\dfrac{4x-3}{x+4} =4 \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to\infty}\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right)\)
Multiply by the conjugate. Then divide by the largest term in the denominator.
\(\displaystyle \lim_{x\to\infty}\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right)=-\infty\)
We multiply by the conjugate: \[\begin{aligned} \lim_{x\to\infty}&\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right)\\ &=\lim_{x\to\infty}\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right) \cdot \dfrac{\sqrt{x+x^2}+\sqrt{5x+3x^2}}{\sqrt{x+x^2}+\sqrt{5x+3x^2}}\\ &=\lim_{x\to\infty}\dfrac{(x+x^2)-(5x+3x^2)}{\sqrt{x+x^2}+\sqrt{5x+3x^2}}\\ &=\lim_{x\to\infty}\dfrac{-4x-2x^2}{\sqrt{x+x^2}+\sqrt{5x+3x^2}} \end{aligned}\] We now divide the numerator and denominator by the largest power of \(x\) in the denominator, which is \(\sqrt{x^2}=x\): \[\begin{aligned} \lim_{x\to\infty}&\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right)\\ &=\lim_{x\to\infty} \dfrac{-4x-2x^2}{\sqrt{x+x^2}+\sqrt{5x+3x^2}}\cdot \dfrac{\dfrac{1}{x}}{\dfrac{1}{\sqrt{x^2}}} \\ &=\lim_{x\to\infty} \dfrac{-4-2x}{\sqrt{\dfrac{1}{x}+1}+\sqrt{\dfrac{5}{x}+3}} =-\infty \end{aligned}\]
tj
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\(\displaystyle \lim_{x\to\infty}\left(x-x\cos\left(\dfrac{1}{x}\right)\right)\)
Replace \(x\) with its reciprocal.
\(\displaystyle \lim_{x\to\infty}\left(x-x\cos\left(\dfrac{1}{x}\right)\right)=0\)
We replace \(x\) with its reciprocal: \[\begin{aligned} \lim_{x\to\infty}\left(x-x\cos\left(\dfrac{1}{x}\right)\right) &=\lim_{t\to0^+}\left(\dfrac{1}{t}-\dfrac{1}{t}\cos\left(t\right)\right)\\ &=\lim_{t\to0^+}\dfrac{1-\cos\left(t\right)}{t} =0 \end{aligned}\] which is a standard trig limit.
tj
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\(\displaystyle \lim_{x\to0}\dfrac{x^3-4x^2+2x+3}{7x^2+5x-1} \)
Use the limit laws.
\(\displaystyle \lim_{x\to0}\dfrac{x^3-4x^2+2x+3}{7x^2+5x-1} = -3\)
We use the limit laws: \[\begin{aligned} \lim_{x\to0}\dfrac{x^3-4x^2+2x+3}{7x^2+5x-1} &=\dfrac{\lim\limits_{x\to0}\left(x^3-4x^2+2x+3\right)}{\lim\limits_{x\to0} \left(7x^2+5x-1\right)} \\ &=\dfrac{0^3-4\cdot0^2+2\cdot0+3}{7\cdot0^2+5\cdot0-1} =-3 \end{aligned}\] The Quotient Rule was valid because the limit of the denominator came out to be \(-1\).
tj
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\(\displaystyle \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}\)
Factor the numerator and cancel.
\(\displaystyle \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}=2\)
We factor the numerator and cancel: \[\begin{aligned} \lim_{\theta\to0}&\dfrac{\sin^2\theta}{1-\cos\theta} =\lim_{\theta\to0}\dfrac{1-\cos^2\theta}{1-\cos\theta}\\ &=\lim_{\theta\to0}\dfrac{ (1+\cos\theta)(1-\cos\theta) }{1-\cos\theta}\\ &=\lim_{\theta\to0}(1+\cos\theta) =2 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}x\sec x\)
Use the product rule.
\(\displaystyle\lim_{x\to0}x\sec x = 0\)
We use the product rule: \[\begin{aligned} \lim_{x\to0}x\sec x &=\lim_{x\to0}x \cdot \lim_{x\to0}\sec x \\ &=0\cdot1 =0 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to3}\dfrac{x^2+x-12}{x-3}\)
Factor the numerator.
\(\displaystyle\lim_{x\to3}\dfrac{x^2+x-12}{x-3}=7\)
We factor the numerator and cancel: \[\begin{aligned} \lim_{x\to3}\dfrac{x^2+x-12}{x-3} &=\lim_{x\to3}\dfrac{(x-3)(x+4)}{x-3} \\ &\qquad\quad=\lim_{x\to3}(x+4) =7 \end{aligned}\]
tj
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\(\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4}\)
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Solve with the Sandwich Theorem.
Start with \(-1 < x < 1 \).
\(\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4}=\dfrac{1}{2}\)
We know that \(x\) goes to \(0\), so we will try this: \[ -1 < x < 1 \] Now, we need to build up the function in the middle: \[ -1 < x < 1 \] \[ 1 < x + 2 < 3 \] \[ x^2 < x^3 + 2x^2 < 3x^2 \] \[ x^2 + 2 < x^3 + 2x^2 +2 < 3x^2 + 2 \] \[ \dfrac{ x^2 + 2 }{ x + 4 } < \dfrac{ x^3 + 2x^2 + 2 }{ x + 4 } < \dfrac{ 3x^2 + 2 }{ x + 4 } \] Next, we need to evaluate \(\displaystyle \lim_{x\to0}\dfrac{ x^2 + 2 }{ x + 4 }\) and \(\displaystyle \lim_{x\to0}\dfrac{ 3x^2 + 2 }{ x + 4 }\): \[\begin{aligned} \lim_{x\to0}\dfrac{ x^2 + 2 }{ x + 4 } &=\dfrac{ \lim\limits_{x\to0}(x^2 + 2) }{ \lim\limits_{x\to0}(x + 4) }\\ &=\dfrac{ 0 + 2 }{ 0 + 4 } =\dfrac{2}{4} =\dfrac{1}{2} \end{aligned}\] \[\begin{aligned} \lim_{x\to0}\dfrac{ 3x^2 + 2 }{ x + 4 } &=\dfrac{ \lim\limits_{x\to0}(3x^2 + 2) }{ \lim\limits_{x\to0}(x + 4) }\\ &=\dfrac{ 0 + 2 }{ 0 + 4 } =\dfrac{ 2 }{ 4 } =\dfrac{ 1 }{ 2 } \end{aligned}\] Since: \[ \lim_{x\to0}\dfrac{ x^2 + 2 }{ x + 4 }=\lim_{x\to0}\dfrac{ 3x^2 + 2 }{ x + 4 } =\dfrac{ 1 }{ 2 } \] we conclude: \[ \lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4} =\dfrac{1}{2} \] also.
tj
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Solve without the Sandwich Theorem.
Use the quotient rule.
\(\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4}=\dfrac{1}{2}\)
We just use the quotient rule: \[\begin{aligned} \lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4} &=\dfrac{\lim\limits_{x\to0}(x^3+2x^2+2)}{\lim\limits_{x\to0}(x+4)}\\ &=\dfrac{\lim\limits_{x\to0}x^3+\lim\limits_{x\to0}2x^2+\lim\limits_{x\to0}2}{\lim\limits_{x\to0}x+\lim\limits_{x\to0}4}\\ &\qquad=\dfrac{0+0+2}{0+4} =\dfrac{2}{4} =\dfrac{1}{2} \end{aligned}\]
tj
We get \(\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2+2}{x+4}=\dfrac{1}{2}\), which matches what we got in part A.
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\(\displaystyle \lim_{x\to\infty} \left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right) \)
Multiply by the conjugate and then divide by the largest term in the denominator.
\(\displaystyle \lim_{x\to\infty} \left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right)=2 \)
We multiply by the conjugate: \[\begin{aligned} \lim&_{x\to\infty} \left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right)\\ &=\lim_{x\to\infty} \left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right) \\ &\qquad\qquad\cdot \dfrac{\sqrt{x^{2}+3x+3}+\sqrt{x^{2}-x+4}}{\sqrt{x^{2}+3x+3}+ \sqrt{x^{2}-x+4}}\\ &=\lim_{x\to\infty} \dfrac{\left(x^{2}+3x+3\right)-\left(x^{2}-x+4\right)} {\sqrt{x^{2}+3x+3}+\sqrt{x^{2}-x+4}} \\ &=\lim_{x\to\infty} \dfrac{4x-1}{\sqrt{x^{2}+3x+3}+\sqrt{x^{2}-x+4}} \end{aligned}\] We now divide the numerator and denominator by the largest power of \(x\) in the denominator, which is \(\sqrt{x^2}=x\): \[\begin{aligned} \lim&_{x\to\infty} \left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right)\\ &=\lim_{x\to\infty} \dfrac{4x-1}{\sqrt{x^{2}+3x+3}+\sqrt{x^{2}-x+4}}\cdot \dfrac{\dfrac{1}{x}}{\dfrac{1}{\sqrt{x^2}}} \\ &=\lim_{x\to\infty} \dfrac{4-\dfrac{1}{x}}{\sqrt{1+\dfrac{3}{x}+\dfrac{3}{x^2}} +\sqrt{1-\dfrac{1}{x}+\dfrac{4}{x^2}}} =\dfrac{4}{2} =2 \end{aligned}\]
tj
PY: Checked to here
Use the Limit Laws to compute each of the following limits. Identify which Limit Laws you used.
Compute each limit. (Identify the trick used.)
Use the Sandwich Theorem to compute each of the following limits.
Use the limit \(\displaystyle \lim_{x\to0}\dfrac{\sin x}{x}=1\) to compute each of the following limits.
Evaluate the limit.
Review Exercises
Solve the limits.
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