4. Computing Limits

Exercises

a. Limits at Finite Numbers

Use the Limit Laws to compute each of the following limits. Identify which Limit Laws you used.

\(\displaystyle\lim_{t\to23}30\)

What is the limit of a constant?

Constant Function Law.
\(\displaystyle\lim_{t\to23}30=30\)

Constant Function Law:
\(30\) is a constant. So: \[ \lim_{t\to23}30 =30 \]

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\(\displaystyle\lim_{x\to2}5^x\)

Use the Exponential Law.

Exponential Law.
\(\displaystyle\lim_{x\to2}5^x=25\)

Exponential Law:
The constant base is \(b=5\). So: \[ \lim_{x\to2}5^x=5^2=25 \]

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\(\displaystyle\lim_{x\to-3}(x+x^2)\)

Use the Addition, Identity and Power Laws.

Addition, Identity and Power Laws.
\(\displaystyle\lim_{x\to-3}(x+x^2)=6\)

We use the Addition Law: \[ \lim_{x\to-3}(x+x^2) =\lim_{x\to-3}x+\lim_{x\to-3}x^2 \] Next, we use the Identity and Power Laws: \[ \lim_{x\to-3}(x+x^2) =-3+(-3)^2=6 \]

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\(\displaystyle\lim_{t\to6}\dfrac{t^2+1}{t}\)

Use the Quotient, Addition, and Power Laws.

Quotient, Addition, Identity, Power and Constant Function Laws.
\(\displaystyle\lim_{t\to6}\dfrac{t^2+1}{t}=\dfrac{37}{6}\)

We use the Quotient Law: \[ \lim_{t\to6}\dfrac{t^2+1}{t} =\dfrac{\lim\limits_{t\to6}(t^2+1)}{\lim\limits_{t\to6}t} \] Notice the Quotient Law is valid because the denominator is \(\lim\limits_{t\to6}t=6\ne0\). Next, we use the Addition and Identity Laws: \[ \lim_{t\to6}\dfrac{t^2+1}{t} =\dfrac{\lim\limits_{t\to6}t^2+\lim\limits_{t\to6}1}{6} \] Finally, we use the Power and Constant Function Laws: \[ \lim_{t\to6}\dfrac{t^2+1}{t} =\dfrac{6^2+1}{6}=\dfrac{37}{6} \]

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\(\displaystyle\lim_{y\to0}\dfrac{1}{y^2+1}\)

Use the Reciprocal and Power Laws.

Reciprocal, Addition, Power and Constant Function Laws.
\(\displaystyle\lim_{y\to0}\dfrac{1}{y^2+1}=1\)

We use the Reciprocal and Addition Laws: \[ \lim_{y\to0}\dfrac{1}{y^2+1} =\dfrac{1}{\lim\limits_{y\to0}(y^2+1)} =\dfrac{1}{\lim\limits_{y\to0}y^2+\lim\limits_{y\to0}1} \] Now, we use the Power and Constant Function Laws: \[ \lim_{y\to0}\dfrac{1}{y^2+1} =\dfrac{1}{0^2+1} =\dfrac{1}{1}=1 \] Notice the Reciprocal Law was valid because the denominator came out to be \(\displaystyle \lim_{y\to0}(y^2+1)=1\ne0\).

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\(\displaystyle\lim_{y\to4}3y\)

Use the Constant Multiple and Identity Laws.

Constant Multiple and Identity Laws.
\(\displaystyle\lim_{y\to4}3y=12\)

We use the Constant Multiple and Identity Laws.: \[\begin{aligned} \lim_{y\to4}3y =3\lim_{y\to4}y =3\cdot4 =12 \end{aligned}\]

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If \(\displaystyle\lim_{x\to3}f(x)=7\) and \(\displaystyle \lim_{x\to3}g(x)=4\), compute: \[\lim_{x\to3} \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2} \]

\(\displaystyle\lim_{x\to3} \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}=\dfrac{17}{15}\)

We start with the Quotient Law: \[\begin{aligned} \lim_{x\to3}& \dfrac{g(x)^2-4f(x)g(x)+3f(x)^2}{-2g(x)^2+f(x)g(x)+f(x)^2}\\ &=\dfrac{\lim\limits_{x\to3}\left(g(x)^2-4f(x)g(x)+3f(x)^2\right)} {\lim\limits_{x\to3}\left(-2g(x)^2+f(x)g(x)+f(x)^2\right)} \end{aligned}\] We use the Addition, Subtraction and Constant Multiple Laws: \[ =\dfrac{\lim\limits_{x\to3}\left[g(x)^2\right] -4\lim\limits_{x\to3}\left[f(x)g(x)\right] +3\lim\limits_{x\to3}\left[f(x)^2\right]} {-2\lim\limits_{x\to3}\left[g(x)^2\right]+ \lim\limits_{x\to3}\left[f(x)g(x)\right] +\lim\limits_{x\to3}\left[f(x)^2\right]} \] Next, we use the Product and Power Laws: \[ =\lim_{x\to3} \dfrac{\left[\lim\limits_{x\to3}g(x)\right]^2 -4\lim\limits_{x\to3}f(x)\lim\limits_{x\to3}g(x) +3\left[\lim\limits_{x\to3}f(x)\right]^2} {-2\left[\lim\limits_{x\to3}g(x)\right]^2 +\lim\limits_{x\to3}f(x)\lim\limits_{x\to3}g(x) +\left[\lim\limits_{x\to3}f(x)\right]^2} \] Finally we plug in the given values: \[ =\lim_{x\to3} \dfrac{\left[4\right]^2 -4\cdot7\cdot4 +3\left[7\right]^2} {-2\left[4\right]^2 +7\cdot4 +\left[7\right]^2} =\dfrac{51}{45}=\dfrac{17}{15} \] Since the limit of the denominator came out to be \(45\) which is non-zero, the Quotient Law used at the start is valid.

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b. When Limit Laws Don't Apply

Compute each limit. (Identify the trick used.)

\(\displaystyle \lim_{x\to2}\dfrac{x^2+2x-8}{x-2}\)

Factor the numerator and cancel.

\(\displaystyle \lim_{x\to2}\dfrac{x^2+2x-8}{x-2}=6\)

We factor the numerator and cancel: \[\begin{aligned} \lim_{x\to2}\dfrac{x^2+2x-8}{x-2} &=\lim_{x\to2}\dfrac{(x-2)(x+4)}{x-2} \\ &=\lim_{x\to2}(x+4)=6 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to-2}\dfrac{x^2-x-6}{x+2} &=\lim_{x\to-2}\dfrac{(x+2)(x-3)}{x+2} \\ &=\lim_{x\to-2}(x-3)=-5 \end{aligned}\]

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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 factor the numerator and cancel: \[\begin{aligned} \lim_{k\to7}\dfrac{k^2-4k-21}{k-7} &=\lim_{k\to7}\dfrac{(k+3)(k-7)}{k-7}\\ &=\lim_{k\to7}(k+3) =10 \end{aligned}\]

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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: \[\begin{aligned} \lim_{k\to-5}\dfrac{k+5}{k^2+2k-15} &=\lim_{k\to-5}\dfrac{k+5}{(k+5)(k-3)}\\ &=\lim_{k\to-5}\dfrac{1}{k-3} =-\,\dfrac{1}{8} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to21}&\dfrac{x^2-441}{x-21} =\lim_{x\to21}\dfrac{(x+21)(x-21)}{x-21}\\ &=\lim_{x\to21}(x+21) =21+21 =42 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0}&\dfrac{(x-4)^2-16}{x} =\lim_{x\to0}\dfrac{x^2-8x}{x} \\ &=\lim_{x\to0}(x-8) =-8 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to3}&\dfrac{(5-x)^2-4}{x-3} =\lim_{x\to3}\dfrac{x^2-10x+21}{x-3} \\ &=\lim_{x\to3}\dfrac{(x-3)(x-7)}{x-3} =\lim_{x\to3}(x-7) =-4 \end{aligned}\]

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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: \[\begin{aligned} \lim_{h\to0}&\dfrac{(4+h)^2-16}{h} =\lim_{h\to0}\dfrac{8h+h^2}{h} \\ &=\lim_{x\to0}(8+h) =8 \end{aligned}\]

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\(\displaystyle \lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}\)

Expand the numerator and cancel. Notice \(x\) is a constant as we compute an \(h\) limit.

\(\displaystyle \lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}=2x\)

We expand the numerator and cancel: \[\begin{aligned} \lim_{h\to0}&\dfrac{(x+h)^2-x^2}{h} =\lim_{h\to0}\dfrac{2hx+h^2}{h} \\ &=\lim_{h\to0}(2x+h) =2x \end{aligned}\]

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We will shortly learn that this is one way to compute the derivative of \(x^2\).
\(\displaystyle \lim_{x\to5} \dfrac{ (x-10)^2 - 25 }{ x - 5 } \)

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: \[\begin{aligned} \lim_{x\to5}&\dfrac{ (x-10)^2 - 25 }{ x - 5 } =\lim_{x\to5} \dfrac{ x^2 - 20x + 100 - 25 }{ x - 5 }\\ &=\lim_{x\to5} \dfrac{ x^2 - 20x + 75 }{ x - 5 } =\lim_{x\to5} \dfrac{ (x - 5)(x - 15) }{ x - 5 }\\ &=\lim_{x\to5} (x-15) =5-15 =-10 \end{aligned}\]

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\(\displaystyle \lim_{x\to0} \left(\dfrac{8x^3}{3}-4x^2\right)\left(\dfrac{2}{x^2}+6\right) \)

This limit has the indeterminate form \(0\cdot\infty\). Try multiplying it out.

\( \lim_{x\to0} \left(\dfrac{8x^3}{3}-4x^2\right)\left(\dfrac{2}{x^2}+6\right) =-8\)

This limit has the indeterminate form \(0\cdot\infty\). We try 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
\(\displaystyle\lim_{x\to0}\dfrac{8x^2-3x}{x^2+5x}\)

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 \(x\): \[\begin{aligned} \lim_{x\to0}&\dfrac{8x^2-3x}{x^2+5x} =\lim_{x\to0}\dfrac{(8x^2-3x)}{(x^2+5x)} \dfrac{\dfrac{1}{x}}{\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{8x-3}{x+5} =-\,\dfrac{3}{5} \end{aligned}\]

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\(\displaystyle\lim_{x\to0}\dfrac{x^2+3x}{2x^3-7x}\)

Divide both the numerator and the denominator by the largest term in the denominator. For \(x\) close to \(0\), which is larger: \(x\) or \(x^3\)?

\(\displaystyle\lim_{x\to0}\dfrac{x^2+3x}{2x^3-7x}=-\,\dfrac{3}{7}\)

For \(x\) close to \(0\), which is larger: \(x\) or \(x^3\)? At \(x=.1\), we have \(x^3=.001\) and so \(x\) is larger. We divide both the numerator and the denominator by the largest term in the denominator: \[\begin{aligned} \lim_{x\to0}\dfrac{x^2+3x}{2x^3-7x} &=\lim_{x\to0}\dfrac{(x^2+3x)\dfrac{1}{x}}{(2x^3-7x)\dfrac{1}{x} }\\ &=\lim_{x\to0}\dfrac{x+3}{2x^2-7} =-\,\dfrac{3}{7} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0}&\dfrac{6x-3x^4}{3x+9x^3} =\lim_{x\to0}\dfrac{(6x-3x^4)\cdot\dfrac{1}{x}}{(3x+9x^3)\cdot\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{6-3x^3}{3+9x^2} =\dfrac{6}{3} =2 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0}\dfrac{4x^2-3x^5}{3x+8x^2} &=\lim_{x\to0}\dfrac{(4x^2-3x^5)\cdot\dfrac{1}{x}}{(3x+8x^2)\cdot\dfrac{1}{x}}\\ &=\lim_{x\to0}\dfrac{4x-3x^4}{3+8x} =\dfrac{0}{3} =0 \end{aligned}\]

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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 \(x\) close 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: \[\begin{aligned} \lim_{x\to0}\dfrac{3x^3-7x^4}{4x^3+x^4} &=\lim_{x\to0}\dfrac{(3x^3-7x^4) \cdot \dfrac{1}{x^3} }{ ( 4x^3+x^4 ) \cdot \dfrac{1}{x^3}}\\ &=\lim_{x\to0}\dfrac{3-7x}{4+x} =\dfrac{3}{4} \end{aligned}\]

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\(\displaystyle \lim_{x\to0} \left( \dfrac{1}{x}-\dfrac{1}{x+x^2+x^3} \right) \)

This limit has the indeterminate form \(\infty-\infty\). Put the terms over a common denominator.

\(\displaystyle \lim_{x\to0}\left(\dfrac{1}{x}-\dfrac{1}{x+x^2+x^3}\right)=1\)

This limit has the indeterminate form \(\infty-\infty\). We put the terms over a common denominator: \[\begin{aligned} \lim_{x\to0} &\left( \dfrac{1}{x}-\dfrac{1}{x+x^2+x^3} \right) \\ &=\lim_{x\to0} \left(\dfrac{1+x+x^2}{x+x^2+x^3}-\dfrac{1}{x+x^2+x^3}\right)\\ &=\lim_{x\to0}\dfrac{x+x^2}{x+x^2+x^3}\cdot\dfrac{\,\dfrac{1}{x}\,}{\dfrac{1}{x}} \\ &=\lim_{x\to0}\dfrac{1+x}{1+x+x^2} =1 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0} &\left(\dfrac{1}{3x+x^2}-\dfrac{1}{3x}\right) =\lim_{x\to0} \left(\dfrac{3}{9x+3x^2}-\dfrac{3+x}{9x+3x^2}\right)\\ &=\lim_{x\to0}\left(\dfrac{-x}{9x+3x^2}\right) =\lim_{x\to0} \left(\dfrac{-1}{9+3x}\right) =-\,\dfrac{1}{9} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0}&\left(\dfrac{1}{x}-\dfrac{1}{x+2x^2}\right) =\lim_{x\to0}\left(\dfrac{1+2x}{x+2x^2}-\dfrac{1}{x+2x^2}\right) \\ &=\lim_{x\to0}\dfrac{2x}{x+2x^2} =\lim_{x\to0}\dfrac{2}{1+2x} =2 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to0}&\left( \dfrac{3+2x^3}{x+4x^3} - \dfrac{x^3+6}{2x+x^3} \right) \\ &=\lim_{x\to0}\left(\dfrac{(3+2x^3)(2x+x^3)-(x^3+6)(x+4x^3)} {(x+4x^3)(2x+x^3)}\right)\\ &=\lim_{x\to0}\left(\dfrac{(6x+3x^3+4x^4+2x^6)-(x^4+4x^6+6x+24x^3)} {2x^2+9x^4+4x^6}\right)\\ &=\lim_{x\to0}\dfrac{-21x^3+3x^4-2x^6}{2x^2+9x^4+4x^6}\cdot \dfrac{\,\dfrac{1}{x^2}\,}{\dfrac{1}{x^2}} \\ &=\lim_{x\to0} \dfrac{-21x+3x^2-2x^4}{2+9x^2+4x^4} =0 \end{aligned}\]

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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 but 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. \[ \lim_{x\to0}\left( \dfrac{8x^2-7x^3}{x^2+3x^3} + \dfrac{9x^2+9x^3}{4x+x^3} \right) =\dfrac{8-7\cdot0}{1+3\cdot0} + \dfrac{9\cdot0+9\cdot0^2}{4+0^2} =8 \]

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It will also work if you put the terms over a common denominator but it is much harder!
\(\displaystyle \lim_{x\to3}\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } \)

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: \[\begin{aligned} \lim_{x\to3}\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } &=\lim_{x\to3}\left(\dfrac{ \sqrt{3x-6} - \sqrt{x} }{ x-3 } \cdot\dfrac{ \sqrt{3x-6} + \sqrt{x} }{ \sqrt{3x-6} + \sqrt{x} }\right)\\ &=\lim_{x\to3}\dfrac{ (3x-6) - (x) }{ (x-3)(\sqrt{3x-6} + \sqrt{x}) }\\ &=\lim_{x\to3}\dfrac{ 2x - 6 }{ (x-3)(\sqrt{3x-6} + \sqrt{x}) }\\ &=\lim_{x\to3}\dfrac{ 2 }{ \sqrt{3x-6} + \sqrt{x} }\\ &=\dfrac{2}{\sqrt{3}+\sqrt{3}} =\dfrac{1}{\sqrt{3}} \end{aligned}\]

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\(\displaystyle \lim_{x\to6}\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6}\)

Multiply by the conjugate of the numerator.

\(\displaystyle \lim_{x\to6}\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6}=\dfrac{1}{2\sqrt{6}}\)

We multiply by the conjugate of the numerator and then multiply out the numerator: \[\begin{aligned} \lim_{x\to6}&\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6} \\ &=\lim_{x\to6}\left(\dfrac{\sqrt{2x-6}-\sqrt{x}}{x-6} \cdot \dfrac{ \sqrt{2x-6}+\sqrt{x} }{ \sqrt{2x-6}+\sqrt{x} } \right) \\ &=\lim_{x\to0}\dfrac{(2x-6)-(x)}{(x-6)(\sqrt{2x-6}+\sqrt{x})}\\ &=\lim_{x\to6}\dfrac{1}{\sqrt{2x-6}+\sqrt{x}} =\dfrac{1}{2\sqrt{6}} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\right) &=\lim_{x\to3}\left(\dfrac{\sqrt{1+x}-\sqrt{7-x}}{x-3}\cdot\dfrac{ \sqrt{1+x}+\sqrt{7-x}}{\sqrt{1+x}+\sqrt{7-x}}\right)\\ &=\lim_{x\to3}\left( \dfrac { (1+x) - (7-x) } { (x-3)( \sqrt{1+x}+ \sqrt{7-x} ) } \right)\\ &=\lim_{x\to3}\left( \dfrac { 2x-6 } { (x-3)( \sqrt{1+x}+ \sqrt{7-x} ) } \right)\\ &=\lim_{x\to3}\left( \dfrac { 2 } { \sqrt{1+x}+ \sqrt{7-x} } \right) =\dfrac{2}{4} =\dfrac{1}{2} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to4}&\dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x} \\ &=\lim_{x\to4} \left( \dfrac{\sqrt{x^2+x}-\sqrt{9x-x^2}}{x^2-4x}\cdot \dfrac{ \sqrt{x^2+x}+\sqrt{9x-x^2} }{ \sqrt{x^2+x}+\sqrt{9x-x^2} } \right)\\ &=\lim_{x\to4} \dfrac{(x^2+x)-(9x-x^2)}{(x^2-4x)(\sqrt{x^2+x}+\sqrt{9x-x^2})}\\ &=\lim_{x\to4} \dfrac{2x^2-8x}{(x^2-4x)(\sqrt{x^2+x}+\sqrt{9x-x^2})}\\ &=\lim_{x\to4} \dfrac{2}{\sqrt{x^2+x}+\sqrt{9x-x^2}} =\dfrac{1}{\sqrt{20}} \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to4}&\dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} }\\ &=\lim_{x\to4} \left( \dfrac{ x-4 }{ \sqrt{x+1} - \sqrt{9-x} } \cdot \dfrac{ \sqrt{x+1} + \sqrt{9-x} }{ \sqrt{x+1} + \sqrt{9-x} }\right)\\ &=\lim_{x\to4} \left( \dfrac{ (x-4)(\sqrt{x+1} + \sqrt{9-x}) }{ (x+1) - (9-x) } \right)\\ &=\lim_{x\to4} \left( \dfrac{ (x-4)(\sqrt{x+1} + \sqrt{9-x}) }{ 2x-8 } \right)\\ &=\lim_{x\to4} \left( \dfrac{ \sqrt{x+1} + \sqrt{9-x} }{ 2 }\right) =\dfrac{\sqrt{5}+\sqrt{5}}{2} =\sqrt{5} \end{aligned}\]

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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: \[\begin{aligned} \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta} &=\lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta}\cdot\dfrac{1+\cos\theta}{1+\cos\theta}\\ &=\lim_{\theta\to0}\dfrac{\sin^2\theta(1+\cos\theta)}{1-\cos^2\theta} \\ &=\lim_{x\to0}(1+\cos\theta) =2 \end{aligned}\]

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As an alternate we can just use a trig identity. Then factor and cancel: \[\begin{aligned} \lim_{\theta\to0}\dfrac{\sin^2\theta}{1-\cos\theta} &=\lim_{\theta\to0}\dfrac{1-\cos^2\theta}{1-\cos\theta}\\ &=\lim_{\theta\to0}\dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta}\\ &=\lim_{x\to0}(1+\cos\theta) =2 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to4}\dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} &=\lim_{x\to4} \left( \dfrac{x-4}{\sqrt{2x-4}-\sqrt{x}} \cdot \dfrac{ \sqrt{2x-4}+\sqrt{x} }{ \sqrt{2x-4}+\sqrt{x} } \right)\\ &=\lim_{x\to4} \left( \dfrac{(x-4)(\sqrt{2x-4}+\sqrt{x})} {(2x-4)-(x)}\right)\\ &=\lim_{x\to4} \left( \dfrac{(x-4)(\sqrt{2x-4}+\sqrt{x})} {x-4}\right)\\ &=\lim_{x\to4} \left(\sqrt{2x-4}+\sqrt{x} \right) =4 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to3}&\dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}} \\ &=\lim_{x\to3} \left( \dfrac{x-3}{\sqrt{2x^2-14}-\sqrt{x^2-5}}\cdot \dfrac{\sqrt{2x^2-14}+\sqrt{x^2-5}}{\sqrt{2x^2-14}+\sqrt{x^2-5}} \right) \\ &=\lim_{x\to3} \dfrac{(x-3)(\sqrt{2x^2-14}+\sqrt{x^2-5})} {(2x^2-14)-(x^2-5)} \\ &=\lim_{x\to3} \dfrac{(x-3)(\sqrt{2x^2-14}+\sqrt{x^2-5})} {x^2-9} \\ &=\lim_{x\to3} \dfrac{\sqrt{2x^2-14}+\sqrt{x^2-5}} {x+3} =\dfrac{4}{6}=\dfrac{2}{3} \end{aligned}\]

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c. Sandwich Theorem

Use the Sandwich Theorem to compute each of the following limits.

\(\displaystyle\lim_{x\to0}x\sin\left(\dfrac{1}{x}\right)\)

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.

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Use the limit \(\displaystyle \lim_{x\to0}\dfrac{\sin x}{x}=1\) to compute each of the following limits.

\(\displaystyle\lim_{\theta\to0}\dfrac{1-\cos^4\theta}{\theta^2}\)

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: \[\begin{aligned} \lim_{\theta\to0}\dfrac{1-\cos^4\theta}{\theta^2} &=\lim_{\theta\to0}\dfrac{ (1+\cos^2\theta) ( 1-\cos^2\theta ) }{\theta^2}\\ &=\lim_{x\to0}( 1+\cos^2\theta ) \cdot \lim_{\theta\to0} \dfrac{ \sin^2\theta }{\theta^2}\\ &=2\cdot(1)^2 =2 \end{aligned}\]

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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: \[\begin{aligned} \lim_{\theta\to0}&\dfrac{1-\sec\theta}{\theta^2} =\lim_{\theta\to0} \left( \dfrac{1-\sec\theta}{\theta^2} \cdot \dfrac{1+\sec\theta}{1+\sec\theta} \right)\\ &=\lim_{\theta\to0}\dfrac{1-\sec^2\theta}{(\theta^2)(1+\sec\theta)} =\lim_{\theta\to0}\dfrac{-\tan^2\theta}{(\theta^2)(1+\sec\theta)}\\ &=-\lim_{\theta\to0}\dfrac{\sin^2\theta}{\theta^2} \lim_{\theta\to0}\dfrac{1}{\cos^2\theta(1+\sec\theta)}\\ &=-\left(\lim_{\theta\to0}\dfrac{\sin\theta}{\theta}\right)^2 \lim_{\theta\to0}\dfrac{1}{\cos^2\theta+\cos\theta} =-\dfrac{1}{2} \end{aligned}\]

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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 \(\displaystyle \lim_{\theta\to0}\dfrac{\tan^2(\theta)}{ \theta }=0\): \[ \lim_{\theta\to0} \sqrt{ \dfrac{ \tan^2(\theta) }{ \theta } +3} =\sqrt{ \left(\lim_{\theta\to0}\dfrac{\tan^2(\theta) }{ \theta }\right)+3} =\sqrt{3} \]

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d. Limits at Infinity

Evaluate the limit.

\(\displaystyle\lim_{x\to\infty}\dfrac{1}{x}\)

Use a Power Law.

\(\displaystyle\lim_{x\to\infty}\dfrac{1}{x}=0\)

This is one of the Power Laws: \[ \lim_{x\to\infty}\dfrac{1}{x} =\lim_{x\to\infty}\dfrac{1}{x^1} =0 \]

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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, we divide by the largest term in the denominator. Here, if \(x\) is large, then \(x\) is larger than \(1\). \[\begin{aligned} \lim_{x\to\infty}&\dfrac{x}{x+1} =\lim_{x\to\infty}\dfrac{x}{x+1} \cdot\dfrac{\dfrac{1}{x}}{\dfrac{1}{x}} \\ &=\lim_{x\to\infty}\dfrac{1}{1+\dfrac{1}{x}} =1 \end{aligned}\]

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\(\displaystyle \lim_{x \to \infty} \left(\dfrac{x^2}{2x+1}-\dfrac{x^2}{2x-2}\right) \)

\(\displaystyle \lim_{x \to \infty} \left(\dfrac{x^2}{2x+1}-\dfrac{x^2}{2x-2}\right) =\dfrac{-3}{4} \)

Plugging in \(x=\infty\) gives the indeterminate form \(\infty-\infty\). We put the fractions using the common denominator: \[\begin{aligned} \lim_{x \to \infty} &\left(\dfrac{x^2}{2x+1}-\dfrac{x^2}{2x-2}\right) = \lim_{x\to\infty} \dfrac{x^2(2x-2)-x^2(2x+1)}{(2x+1)(2x-2)} \\ &= \lim_{x\to\infty} \dfrac{-3x^2}{4x^2-2x-2} =-\dfrac{3}{4} \end{aligned}\]
\(\displaystyle \lim_{x\to\infty}\left(x-\dfrac{x^2+3}{x+4}\right) \)

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: \[\begin{aligned} \lim_{x\to\infty}\left(x-\dfrac{x^2+3}{x+4}\right) &=\lim_{x\to\infty}\left(\dfrac{x^2+4x}{x+4}-\dfrac{x^2+3}{x+4}\right)\\ &=\lim_{x\to\infty}\dfrac{4x-3}{x+4} =4 \end{aligned}\]

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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 \(\sqrt{x^2}=x\): \[\begin{aligned} \lim_{x\to\infty}&\left(\sqrt{x+x^2}-\sqrt{5x+3x^2}\right)\\ &=\lim_{x\to\infty} \dfrac{-4x-2x^2}{\sqrt{x+x^2}+\sqrt{5x+3x^2}}\cdot \dfrac{\dfrac{1}{x}}{\dfrac{1}{\sqrt{x^2}}} \\ &=\lim_{x\to\infty} \dfrac{-4-2x}{\sqrt{\dfrac{1}{x}+1}+\sqrt{\dfrac{5}{x}+3}} =-\infty \end{aligned}\]

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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.

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Review Exercises

Compute the limits.

\(\displaystyle \lim_{x\to0}\dfrac{x^3-4x^2+2x+3}{7x^2+5x-1} \)

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 \(-1\).

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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: \[\begin{aligned} \lim_{x\to0}x\sec x &=\lim_{x\to0}x \cdot \lim_{x\to0}\sec x \\ &=0\cdot1 =0 \end{aligned}\]

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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: \[\begin{aligned} \lim_{x\to3}\dfrac{x^2+x-12}{x-3} &=\lim_{x\to3}\dfrac{(x-3)(x+4)}{x-3} \\ &\qquad\quad=\lim_{x\to3}(x+4) =7 \end{aligned}\]

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\(\displaystyle\lim_{x\to\infty}\dfrac{x^2+\sin x}{2x^2+4}\)

Solve with the Sandwich Theorem., starting with \(-1 < \sin x < 1 \).

\(\displaystyle \lim_{x\to\infty}\dfrac{x^2+\sin x}{2x^2+4}=\dfrac{1}{2}\)

For all values of \(x\), we have: \[ -1 < \sin x < 1 \] Now, we need to build up the function in the middle: \[ x^2-1 < x^2+\sin x < x^2+1 \] \[ \dfrac{x^2-1}{2x^2+4} < \dfrac{x^2+\sin x}{2x^2+4} < \dfrac{x^2+1}{2x^2+4} \] However, \[ \lim_{x\to\infty}\dfrac{x^2\pm1}{2x^2+4} =\lim_{x\to\infty}\dfrac{1\pm\dfrac{1}{x^2}}{2+\dfrac{4}{x^2}} =\dfrac{1}{2} \] By the Sandwich Theorem, we conclude: \[ \lim_{x\to\infty}\dfrac{x^2+\sin x}{2x^2+4} =\dfrac{1}{2} \] also.

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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 \(\sqrt{x^2}=x\): \[\begin{aligned} \lim_{x\to\infty} &\left(\sqrt{x^{2}+3x+3}-\sqrt{x^{2}-x+4}\right)\\ &=\lim_{x\to\infty} \dfrac{4x-1}{\sqrt{x^{2}+3x+3}+\sqrt{x^{2}-x+4}}\cdot \dfrac{\dfrac{1}{x}}{\dfrac{1}{\sqrt{x^2}}} \\ &=\lim_{x\to\infty} \dfrac{4-\dfrac{1}{x}}{\sqrt{1+\dfrac{3}{x}+\dfrac{3}{x^2}} +\sqrt{1-\dfrac{1}{x}+\dfrac{4}{x^2}}} =\dfrac{4}{2} =2 \end{aligned}\]

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