I guess that's an easy one about limits.... Which one of the following is $$\lim_{x\to 0}\frac{e^x + 1}{e^x- 1}\;,$$
if it exists?
(a) Does not exist
(b) $0$
(c) $\infty$
(d) $1$
I know the answer is a, but why? if i susbtitute $0$ into the function I get $1/0$, so an indetrmination, so shouldnt the result be just either $+\infty$ or $-\infty$? I read something about aproaching  the limit from both the right and the left and see if the results are similar then the limit exist, so I did approach the limit with $-0.5$ and I got $-4$ and from $0.5$ and I got $-4.3$, so the results suggest that the limit is continous, so why the solution is a? 
 A: Suppose that $x$ is very close to $0$, but positive; then the numerator $e^x+1$ is very close to $2$, and the denominator $e^x-1$ is a very small positive number, so the fraction is very large. As $x$ gets closer and closer to $0$ on the positive side, the fraction gets bigger without bound:
$$\lim_{x\to 0^+}\frac{e^x + 1}{e^x- 1}=+\infty\;.$$
Now suppose that $x$ is very close to $0$ but negative. The numerator is still very close to $2$, and the denominator is still very small in absolute value, but now the denominator is negative, and the fraction itself is therefore negative. As $x$ gets closer and closer to $0$ on the negative side, the absolute value of the fraction blows up, just as it did when $x\to 0^+$, but now it’s always negative:
$$\lim_{x\to 0^-}\frac{e^x + 1}{e^x- 1}=-\infty\;.$$
Since the two one-sided limits are different, the two-sided limit does not exist even in the extended real numbers (i.e., including the values $+\infty$ and $-\infty$).
If you graph the function $$f(x)=\frac{e^x+1}{e^x-1}\;,$$ you’ll see that the graph has a vertical asymptote at $x=0$, and that the function gets ‘torn apart’ at the asymptote: on the lefthand side the graph dives down towards $-\infty$, and on the righthand side it climbs towards $+\infty$.
A: A substitution might make this look more familiar: write $u=e^x-1$. When $x\to 0, u\to 0$. So this is
$$\lim_{u\to 0}\frac{u+2}{u}$$
$$\lim_{u\to 0}1+\frac 2u$$
And so you can see that the function has this behaviour precisely because $\frac 1x$ does. When $u$ is negative (meaning $x$ is negative) the function decreases dramatically. When $u$ and $x$ are slightly positive, the function increases to very high values.
A: As $x \to 0$, the numerator approaches 2 and the denominator approaches 0, and the form $2/0$ converges to $+\infty$ when $x \to 0^+$ and to $-\infty$ when $x \to 0^-$. Hence, the 2-sided limit does not exist.
