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?

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Try the values $0.01$ and $-0.01$. One of your problems is that $0.5$ and $-0.5$ are not very close to $0$. Also, it would be nice (and you would get better answers) if you put time into editing your question. As it stands, it does not look like you spent much time into typing up the question, let alone thinking about the mathematics behind it. – JavaMan Feb 13 '13 at 5:56
By the way, when $x=0.5$ the value of $(e^x+1)/(e^x-1)$ is about $4.083$, not $-4.3$. – Rahul Feb 13 '13 at 6:04
The value you mention for $x=0.5$, presumably obtained from a calculator, is wrong. It is much easier (and more accurate) without a calculator. If $x\gt 0$, then $e^x\gt 1$ (remember the graph). But if $x$ is positive and very near $0$, then $e^x$ is very near to $1$, so $\frac{e^x+1}{e^x-1}$ is huge positive. Similarly, if $x$ is close to $0$ but less than $0$, our function is huge negative. – André Nicolas Feb 13 '13 at 7:19

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 $+\infty$.

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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.

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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.

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(the form doesn't converge to $+\infty$; you mean the limit is of the form $2/0^+$, which is equal to $+\infty$) – Hurkyl Feb 13 '13 at 7:52
@Hurkyl yes, thank you. – gt6989b Feb 13 '13 at 17:13