7. Computing Limits
b. When Limit Laws Don't Apply
Limits without Laws
1. Limit Tricks
a. Factor and Cancel
This trick says to factor the numerator and the denominator and then cancel any common factors. It is most appropriate for the indeterminate form \(\dfrac{0}{0}\) when the numerator and denominator are .
A generalized polynomial is a sum of terms each of which is a coefficient times the variable to a numerical power, not necessarily non-negative integers. If all the powers are non-negative integers, then it is a polynomial.
Compute \(\displaystyle \lim_{x\to3}\dfrac{x^2-8x+15}{x-3}\).
If we plug \(x=3\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we factor the numerator and cancel: \[\begin{aligned} \lim_{x\to3}\dfrac{x^2-8x+15}{x-3} &=\lim_{x\to3}\dfrac{(x-3)(x-5)}{x-3} \\ &=\lim_{x\to3}(x-5) =-2 \end{aligned}\]
Compute \(\displaystyle \lim_{x\to4}\dfrac{x^2-5x+4}{x^{3/2}-4x^{1/2}}\).
If we plug \(x=4\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we factor the numerator and denominator and cancel: \[\begin{aligned} \lim_{x\to4}\dfrac{x^2-5x+4}{x^{3/2}-4x^{1/2}} &=\lim_{x\to4}\dfrac{(x-4)(x-1)}{\sqrt{x}(x-4)} \\ &=\lim_{x\to4}\dfrac{x-1}{\sqrt{x}} =\dfrac{3}{2} \end{aligned}\]
Compute \(\displaystyle \lim_{x\to3}\dfrac{x-3}{x^3-27}\).
Recall: \[\begin{aligned} (a-b)(a^2+ab+b^2)&=(a^3+a^2b+ab^2)-(a^2b+ab^2+b^3) \\ &=a^3-b^3 \end{aligned}\]
\(\displaystyle \lim_{x\to3}\dfrac{x-3}{x^3-27}=\dfrac{1}{27}\)
If we plug \(x=3\) into the numerator and denominator, we see this has the indeterminate form \(\dfrac{0}{0}\). To simplify this, we factor the denominator and cancel: \[\begin{aligned} \lim_{x\to3}\dfrac{x-3}{x^3-27} &=\lim_{x\to3}\dfrac{x-3}{(x-3)(x^2+3x+9)} \\ &=\lim_{x\to3}\dfrac{1}{x^2+3x+9} =\dfrac{1}{27} \end{aligned}\]
Compute \(\displaystyle \lim_{x\to25}\dfrac{\sqrt{x}-5}{x-25}\).
Recall: \[ a^2-b^2=(a-b)(a+b) \] This works even if \(a\) and/or \(b\) is a square root.
\(\displaystyle \lim_{x\to25}\dfrac{\sqrt{x}-5}{x-25}=\dfrac{1}{10}\)
Notice: \[ x-25=(\sqrt{x}-5)(\sqrt{x}+5) \] So: \[\begin{aligned} \lim_{x\to25}\dfrac{\sqrt{x}-5}{x-25} &\lim_{x\to25}\dfrac{\sqrt{x}-5}{(\sqrt{x}-5)(\sqrt{x}+5)} \\ &=\lim_{x\to25}\dfrac{1}{\sqrt{x}+5} =\dfrac{1}{10} \end{aligned}\]