Limits of the form 0/0

When both the numerator and the denominator approach $0$ as $x\to a$, the numerator has $x-a$ as a factor (factor theorem). Factoring and canceling leaves an expression you can substitute into.

For cubics, use $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^3+a^3=(x+a)(x^2-ax+a^2)$.

Examples

1. 01What is the value of the limit?$\displaystyle \lim_{x\to -3}\frac{x^2+x-6}{x+3}$AnswerHide

   Answer$\displaystyle -5$

   Steps

   1. $\displaystyle \frac{x^{2}+x-6}{x+3}$$\displaystyle = \frac{(x+3)(x-2)}{x+3}$$\displaystyle = x-2$
   2. Put $-3$ in
   3. $\displaystyle (-3)-2$$\displaystyle = -5$
2. 02What is the value of the limit?$\displaystyle \lim_{x\to -3}\frac{x^2+4x+3}{x+3}$AnswerHide

   Answer$\displaystyle -2$

   Steps

   1. $\displaystyle \frac{x^{2}+4x+3}{x+3}$$\displaystyle = \frac{(x+3)(x+1)}{x+3}$$\displaystyle = x+1$
   2. Put $-3$ in
   3. $\displaystyle (-3)+1$$\displaystyle = -2$
3. 03What is the value of the limit?$\displaystyle \lim_{x\to 2}\frac{x^3-8}{x-2}$AnswerHide

   Answer$\displaystyle 12$

   Steps

   1. $\displaystyle \frac{x^{3}-8}{x-2}$$\displaystyle = \frac{(x-2)(x^{2}+2x+4)}{x-2}$$\displaystyle = x^{2}+2x+4$
   2. Put $2$ in
   3. $\displaystyle 2^{2}+2\cdot2+4$$\displaystyle = 12$
4. 04What is the value of the limit?$\displaystyle \lim_{x\to 3}\frac{x^2-9}{x-3}$AnswerHide

   Answer$\displaystyle 6$

   Steps

   1. $\displaystyle \frac{x^{2}-9}{x-3}$$\displaystyle = \frac{(x-3)(x+3)}{x-3}$$\displaystyle = x+3$
   2. Put $3$ in
   3. $\displaystyle 3+3$$\displaystyle = 6$
5. 05What is the value of the limit?$\displaystyle \lim_{x\to 3}\frac{x^2-5x+6}{x-3}$AnswerHide

   Answer$\displaystyle 1$

   Steps

   1. $\displaystyle \frac{x^{2}-5x+6}{x-3}$$\displaystyle = \frac{(x-3)(x-2)}{x-3}$$\displaystyle = x-2$
   2. Put $3$ in
   3. $\displaystyle 3-2$$\displaystyle = 1$
6. 06What is the value of the limit?$\displaystyle \lim_{x\to -3}\frac{x^3+27}{x+3}$AnswerHide

   Answer$\displaystyle 27$

   Steps

   1. $\displaystyle \frac{x^{3}+27}{x+3}$$\displaystyle = \frac{(x+3)(x^{2}-3x+9)}{x+3}$$\displaystyle = x^{2}-3x+9$
   2. Put $-3$ in
   3. $\displaystyle (-3)^{2}-3\cdot(-3)+9$$\displaystyle = 27$

FAQ

Q1Can I divide by $x-a$ even though it is $0$ at $x=a$?AnswerHide

A

Yes. In the limit $x\to a$, $x$ only approaches $a$ and never equals it, so $x-a\ne0$ and canceling is allowed.

Q2What do I do after canceling?AnswerHide

A

What remains is a polynomial, so the limit is its value at $x=a$. For example, $\frac{x^3-8}{x-2}=x^2+2x+4$, so the limit as $x\to2$ is $4+4+4=12$.