# Finding the limit of an expression having a zero in the denominator

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....

-
did you just edit my post? – codenamejupiterx Feb 14 '13 at 6:27

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.

-

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.

-
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.

-
Ok Thanks!!!!!! – codenamejupiterx Feb 14 '13 at 6:17