Limits by Factoring

Finding a limit by factoring is a technique to finding limits that works by canceling out common factors. This sometimes allows us to transform an indeterminate form into one that allows for direct evaluation.

Use Case Example

Find

$\lim_{x \rightarrow 3}\dfrac{x^2-9}{x-3}.$

Using the substitution rule gives

$\lim_{x \rightarrow 3}\dfrac{x^2-9}{x-3}=\frac{3^2-9}{3-3}=\frac{0}{0},$

which is an indeterminate number. Limits that end in an indeterminate number like the example above are said to be in indeterminate form. In calculus, there are 7 primary indeterminate numbers:

$\begin{array}{c}&\frac{0}{0}, &\frac{\infty}{\infty}, &\infty-\infty, &\infty \cdot 0, &0^0, &\infty^0, &1^\infty.\end{array}$

Then, how do we find the values of limits in indeterminate form? Limits that end in the form of $\frac{0}{0}$ usually can be solved by factoring the numerator and denominator. Then find the common divisor and divide both the numerator and denominator by it. Hence the solution for the example above will be as follows:

$\lim_{x \rightarrow 3}\dfrac{x^2-9}{x-3}=\lim_{x \rightarrow 3}\dfrac{(x-3)(x+3)}{x-3}=\displaystyle{\lim_{x \rightarrow 3}(x+3)}=3+3=6.$

Observe that $\displaystyle{\lim_{x \rightarrow 3}(x-3)}=0$ , and thus the common divisor $x-3$ was the culprit that made the numerator and denominator zero. $_\square$

Worked Examples

What is ${\displaystyle\lim_{x \rightarrow 5}}\frac{x^2-x-20}{x-5}?$

We have

$\begin{aligned} \lim_{x \rightarrow 5}\frac{x^2-x-20}{x-5} &=\lim_{x \rightarrow 5}\frac{(x-5)(x+4)}{x-5}\\\\ &=\lim_{x \rightarrow 5}(x+4)\\\\ &=9. \ _\square \end{aligned}$

What is ${\displaystyle\lim_{h \rightarrow 0}}\frac{(h+4)^2-16}{h}?$

We have

$\begin{aligned} \lim_{h \rightarrow 0}\frac{(h+4)^2-16}{h} &=\lim_{h \rightarrow 0}\frac{h^2+8h+16-16}{h}\\\\ &=\lim_{h \rightarrow 0}\frac{h(h+8)}{h}\\\\ &=\displaystyle{\lim_{h\rightarrow0}(h+8)}\\\\ &=8. \ _\square \end{aligned}$

When we encounter limits with square roots, multiplying the numerator and denominator by the conjugate followed by factoring is usually the solution.

Find $\displaystyle{\lim_{x \rightarrow \infty}\left(\sqrt{x^2+4x}-x\right)}.$

Plugging in $x=\infty$ gives the indeterminate form of $\infty-\infty$ . Using the conjugate $\sqrt{x^2+4x}+x$ gives

$\begin{aligned} \displaystyle{\lim_{x \rightarrow \infty}\left(\sqrt{x^2+4x}-x\right)} &=\lim_{x \rightarrow \infty}\frac{\left(\sqrt{x^2+4x}-x\right)\left(\sqrt{x^2+4x}+x\right)}{\sqrt{x^2+4x}+x}\\\\ &=\lim_{x \rightarrow \infty}\frac{x^2+4x-x^2}{\sqrt{x^2+4x}+x}\\\\ &=\lim_{x \rightarrow \infty}\frac{4x}{\sqrt{x^2+4x}+x}\\\\ &=\lim_{x \rightarrow \infty}\frac{4}{\sqrt{1+\frac{4}{x}}+1}\\\\ &=\frac{4}{2}\\\\ &=2. \ _\square \end{aligned}$

What is ${\displaystyle\lim_{x \rightarrow 0}}\frac{\sqrt{4+3x}-2}{x}?$

This time, using the substitution rule gives $\frac{0}{0}$ . Again, we multiply the numerator and denominator by the conjugate $\sqrt{4+3x}+2$ and get

$\begin{aligned} \lim_{x \rightarrow 0}\frac{\sqrt{4+3x}-2}{x} &=\lim_{x \rightarrow 0}\frac{\left(\sqrt{4+3x}-2\right)\left(\sqrt{4+3x}+2\right)}{x\cdot \left(\sqrt{4+3x}+2\right)}\\\\ &=\lim_{x \rightarrow 0}\frac{(4+3x)-4}{x\cdot \left(\sqrt{4+3x}+2\right)}\\\\ &=\lim_{x \rightarrow 0}\frac{3x}{x\cdot \left(\sqrt{4+3x}+2\right)}\\\\ &=\lim_{x \rightarrow 0}\frac{3}{\sqrt{4+3x}+2}\\\\ &=\frac{3}{4}. \ _\square \end{aligned}$

What is $\displaystyle{\lim_{x \rightarrow \infty}\left(x-\sqrt{x^2-7x+2}\right)}.$

We have $\begin{aligned} \lim_{x \rightarrow \infty}\left(x-\sqrt{x^2-7x+2}\right) &=\lim_{x \rightarrow \infty}\frac{\left(x-\sqrt{x^2-7x+2}\right)\left(x+\sqrt{x^2-7x+2}\right)}{x+\sqrt{x^2-7x+2}}\\\\ &=\lim_{x \rightarrow \infty}\frac{x^2-(x^2-7x+2)}{x+\sqrt{x^2-7x+2}}\\\\ &=\lim_{x \rightarrow \infty}\frac{7x-2}{x+\sqrt{x^2-7x+2}}\\\\ &=\lim_{x \rightarrow \infty}\frac{7-\frac{2}{x}}{1+\sqrt{1-\frac{7}{x}+\frac{2}{x^2}}}\\\\ &=\frac{7}{2}. \ _\square \end{aligned}$

Find $\displaystyle{\lim_{x \rightarrow 8}\frac{\sqrt[3]{x}-2}{x-8}}.$

We have $\begin{aligned} \displaystyle{\lim_{x \rightarrow 8}\frac{\sqrt[3]{x}-2}{x-8}} &=\displaystyle{\lim_{x \rightarrow 8}\frac{\sqrt[3]{x}-2}{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}}\\\\ &=\displaystyle{\lim_{x \rightarrow 8}\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}}\\\\ &=\frac{1}{12}.\ _\square \end{aligned}$

What is $\displaystyle{\lim_{x \rightarrow 0}\frac{1}{x}\left(1-\frac{4}{(x+2)^2}\right)}.$

We have $\begin{aligned} \lim_{x \rightarrow 0}\frac{1}{x}\left(1-\frac{4}{(x+2)^2}\right) &=\lim_{x \rightarrow 0}\frac{1}{x}\cdot \frac{(x+2)^2-4}{(x+2)^2}\\\\ &=\lim_{x \rightarrow 0} \frac{x^2+4x}{x(x+2)^2}\\\\ &=\lim_{x \rightarrow 0} \frac{x+4}{(x+2)^2}\\\\ &=1. \ _\square \end{aligned}$

Find $\displaystyle{\lim_{h \rightarrow 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h}}.$

We have $\begin{aligned} \lim_{h \rightarrow 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h} &=\lim_{h \rightarrow 0}\frac{\hspace{3mm} \frac{-h}{x(x+h)}\hspace{3mm} }{h}\\\\ &=\lim_{h \rightarrow 0}\frac{-1}{x(x+h)}\\\\ &=-\frac{1}{x^2}.\ _\square \end{aligned}$

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