$$ \ \lim_{x \to 0}\frac{e^x + e^{-x} }{x} $$
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 ?
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$\begingroup$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), $$\lim_{x \to 0} \frac{\text{sin}(x)}{x} = \lim_{x \to 0} \frac{\text{cos}(x)}{1} = 1 \,.$$
$\endgroup$ 2 $\begingroup$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.
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