# Does $\lim_{x \to 2} \frac{2x+4}{x^2-4}$ exist?

So the lecturer's assistant is saying that the following limit exists and that it is $\infty$

So the equation $$\lim_{x \to 2} \frac{2x+4}{x^2-4}$$

Now, if I go ahead and simplify the expression first I end up with

$$\lim_{x \to 2} \frac{2}{x-2}$$

and as far as I can tell this limit does not exist. Reason being that the Left Hand Side limit will be $-\infty$ and the Right Hand Side limit will be $+\infty$.

Can someone please verify if I'm in the wrong here?

Much appreciated.

• It should simplify to $2/(x-2)$, and yes, the limit does not exist. Oct 22, 2013 at 7:41
• Thank you, I'll update the simplified answer in question. Oct 22, 2013 at 7:55

$$\lim_{x\to 2}{\frac{2x+4}{x^2-4}} = \lim_{x \to 2}{\frac{2(x+2)}{(x-2)(x+2)}}$$ $$\lim_{x\to 2}{\frac{2}{x-2}}$$ $$\text{Left-hand limit} = \lim_{x\to 2^-}{\frac{2}{x-2}} = -\infty$$ $$\text{Right-hand limit} = \lim_{x\to2^+}{\frac{2}{x-2}}= +\infty$$ $$\text{Left-hand limit} \ne \text{Right-hand limit}$$ Hence limit doesn't exist.
\begin{align} &\lim_{x\to x_0}f(x)=+\infty \\ &\lim_{x\to x_0}f(x)=-\infty \\ &\lim_{x\to x_0}f(x)=\infty \end{align} where the last means $\lim_{x\to x_0}|f(x)|=+\infty$. According to this convention, one can say that $\lim_{x\to0}1/x=\infty$, which is essentially the same as your case.