Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question about undefined rational expressions in calculus with zeros in the denominator. Ok, how is $$\lim_{x\to 2}$$ for some expression that had $$(x-2)$$ in the denominator undefined when the notation $$\lim_{x\to 2}$$ implies $x\not= 2$? Thank You in advance....

share|cite|improve this question
did you just edit my post? – codenamejupiterx Feb 14 '13 at 6:27
up vote 0 down vote accepted

The way these sort of questions crop up is when you're trying to compute $\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ where $\displaystyle\lim_{x\rightarrow a}g(x)=0$.

Of course, if $\displaystyle\lim_{x\rightarrow a}f(x)=0$ as well, we can either reduce the expression $f(x)/g(x)$ or apply L'Hospital's rule.

You're probably more interested in what to do when $\displaystyle\lim_{x\rightarrow a}f(x)\neq0$. When this is the case, the limit $\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$ is either $\pm\infty$ or it does not exist. The way to determine the limit is by looking at the left and right-hand limits $\displaystyle\lim_{x\rightarrow a^-}\frac{f(x)}{g(x)}$ and $\displaystyle\lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}$.

Take $\displaystyle\lim_{x\rightarrow 2}\frac{1}{x-2}$ for example. When $x<2$ the expression $\displaystyle\frac{1}{x-2}$ is negative so we see that $\displaystyle\lim_{x\rightarrow 2^-}\frac{1}{x-2}=-\infty$. However, when $x>2$ the expression $\displaystyle\frac{1}{x-2}$ is positive so $\displaystyle\lim_{x\rightarrow 2^+}\frac{1}{x-2}=\infty$. The left-hand and right-hand limits do not agree, so the limit does not exist! This is apparant from the graph of $\displaystyle\frac{1}{x-2}$.

If we were to alter the problem by looking instead at $\displaystyle\lim_{x\rightarrow 2}\frac{1}{\left|x-2\right|}$ we would see that both the left-hand and right-hand limits are $\infty$, so the original limit is infinity.

share|cite|improve this answer

The notation $\lim_{x\to2}$does not mean "$x$ could equal $2.1$ or $1.9$."

The notation $\lim_{x\to2}f(x)=L$ means for every positive $\epsilon$ there is a number $\delta$ such that if $|x-2|\lt\delta$ then $|f(x)-L|\lt\epsilon$.

If you are not careful with definitions --- if you don't state them properly, and understand them fully --- you are up the creek without a paddle.

share|cite|improve this answer
This answer referred to something in the original version of the question, before the edit by Zilliput. – Gerry Myerson Aug 1 '14 at 3:49

If $$L=\lim_{x\to 2}f(x),$$ where $f(x)$ has $(x-2)$ as a factor in the denominator, then $L$ may or may not be defined. For example, $$L_1=\lim_{x\to 2}\frac{x^2-4}{x-2}=\lim_{x\to 2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\to 2}(x+2)=4.$$ However, $$L_2=\lim_{x\to 2}\frac{1}{x-2}$$ is undefined.

share|cite|improve this answer
Ok Thanks!!!!!! – codenamejupiterx Feb 14 '13 at 6:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.