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$$ \ \lim_{x \to 0}\frac{e^x + e^{-x} }{x} $$

is the problem.

It does not exist.

But if I break up the problem and apply LHospital's to one part ( is it okay to apply the rule to part of the problem ) , I get a finite answer.

For example, $$ \ \lim_{ x \to 0}\frac{e^x + 1 + e^{-x} -1 }{x} $$ which is equal to $$ \ \lim_{ x \to 0}\ 1 + \frac{ e^{-x} +1 }{x} $$ So applying Lhospitals' to only the second part $$ \ \lim_{ x \to 0}\ \frac{ e^{-x} +1 }{x} $$ We get $$ \ \lim_{ x \to 0}\ \frac{-e^{-x}}{1} - \frac{0}{1} = -1 $$

So adding $$1 + (-1) = 0 $$ is the answer . So how can the limit not exist. Same procedure can be done to $$ \ \lim_{ x \to 0}\frac{\cos(x)}{x} $$ , applying LHospitals', you get $$ \ \lim_{x \to 0}\frac{-\sin(x)}{1} = 0 $$ But $$ \ \lim_{ x \to 0}\frac{\cos(x)}{x} $$ also is non-existent. Can someone please explain this to me ?

Curious
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    You're using $Lim (A+B)=LimA+LimB$ . This is only guaranteed when $Lim(A+B)$ exists. – user99680 May 18 '14 at 08:07
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    Why do you replace $\lim\limits_{x\to0}\frac{e^x+1}x$ by $1$? – Did May 18 '14 at 08:08
  • @Did That was a mistake. I mixed it up with the standard integral – Curious May 19 '14 at 10:44
  • At the risk of being insistent and contrary to what is said below, the crux of your problem appears to be the fact that you thought that $\lim\limits_{x\to0}\frac{e^x+1}x$ exists (and that it is $1$). It does not (exist). – Did May 19 '14 at 10:51
  • @Did I've edited the equation so that now you get $$ \ \lim_{ x \to 0}\ \frac{ e^{-x} +1 }{x} $$ which is $$\frac{2}{0} $$ which I thought to be indeterminate and hence applied LHospital's rule. – Curious May 19 '14 at 10:55
  • Sorry but your last edit only made a mess of the question. – Did May 19 '14 at 11:00
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    You only apply L'Hopital's rule to $\lim a/b$ if limits of $a$ and $b$ are both $0$ or both $\infty$. If one is zero and the other is finite and not zero, the limit is known (and L'Hopital's rule is not valid of course) – Jean-Claude Arbaut May 19 '14 at 11:08
  • @Did I applied the standard limit $$ \ \lim_{ x \to 0}\ \frac{ e^{x} - 1 }{x} = 1 $$ , so that the other part is $$ \ \lim_{ x \to 0}\ \frac{ e^{-x} +1 }{x} $$ which was added onto it. But anyway the method was wrong. – Curious May 19 '14 at 11:13
  • Why is it that $$ \frac{n}{0} = \infty \ $$ ? How do we perceive this intuitively ? – Curious May 19 '14 at 11:17
  • After the (strange) modifications you brought to the question, the problem is that $$\lim\limits_{ x \to 0}\frac{ e^{-x} +1 }{x}$$ does not exist. Anyway, I think I will stop here since the reasons (to which I pointed you from my first comment on) why there is a problem (in the previous version of the question as well as in the new one) are on the page and anybody interested can read them. – Did May 19 '14 at 15:02
  • @Did My idea about limits was wrong. – Curious May 19 '14 at 17:04

2 Answers2

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Observe that (consider the behavior of the fraction for the denominator, since the numerator approaches $2$ in either case)$$\begin{align} \lim_{x \to 0^+} \frac{e^x + e^{-x}}{x} = \infty \end{align}\,,$$ but on the other hand, $$\begin{align} \lim_{x \to 0^-} \frac{e^x + e^{-x}}{x} = -\infty \end{align}\,.$$ It follows that this limit does not exist. It is similar in the other case you mentioned as well. We have $$\lim_{x \to 0^+} \frac{\text{cos}(x)}{x} = \infty \\\\ \lim_{x \to 0^-} \frac{\text{cos}(x)}{x} = -\infty \,.$$ Thus, the limit does not exist in this case either. (To clarify notation, $\lim_{x \to 0^+}$ means the limit as $x$ approaches $0$ from the right, while $\lim_{x \to 0^-}$ means the limit as $x$ approaches $0$ from the left. These are called one-sided limits and both must exist and be equivalent for the limit at $0$ to exist.) L'Hôpital's rule cannot apply here: I suggest reading up on the rule and becoming familiar with when exactly you can use it. The fractional limit must fall into a specific list of indeterminate forms. As an example, take the following limit (which has the form $0/0$, allowable by L'Hôpital, and is very similar to the former limit), $$\lim_{x \to 0} \frac{\text{sin}(x)}{x} = \lim_{x \to 0} \frac{\text{cos}(x)}{1} = 1 \,.$$

afedder
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  • The problem was I took $$\frac{1}{0} $$ also as indeterminate and applied LHospital's rule. – Curious May 19 '14 at 10:36
  • Yes, that is the crux of the problem, as well as some other illegal algebraic moves with limits. Here are the only allowable indeterminate forms by L'Hôpital's rule: $\frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$ – afedder May 19 '14 at 10:41
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It is only valid to use the rule $$ \lim_{x \to a}(f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) $$ when you know that all limits exist and are finite.

user139388
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