Finding the limit $ \lim_{x\to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)} $ I have been working on this question for a while now:
$$
\lim_{x\to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}
$$
I know the answer is $\cos a$, that's what the solutions say. But I don't understand how it became to that. I've tried solving it using L'Hopital rule but I've gone nowhere because the expressions just keep getting messier and messier. What's a sensible way to go about doing this? Thank you in advance for your answer.
 A: Note that we can rewrite this as
$$
\lim_{x\to a^+} \cos(x) \cdot \lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} = 
\cos(a)\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)}
$$
In order to show that the remaining limit becomes $1$, use L'Hôpital's rule:
$$
\begin{align}
\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} &=
\lim_{x\to a^+} \frac{1/(x-a)}{1/(e^{x} - e^a)\cdot e^x}\\
&=\lim_{x\to a^+} e^{-x}
\lim_{x\to a^+} \frac{e^x - e^a}{x-a}\\
&= e^{-a} \lim_{x\to a^+} \frac{e^x - e^a}{x-a}
\end{align}
$$
Using L'Hôpital's rule once again gives us
$$
\begin{align}
\lim_{x\to a^+} \frac{\ln(x-a)}{\ln(e^{x} - e^a)} &=
e^{-a} \lim_{x\to a^+} \frac{e^x}{1} = e^{-a}e^a = 1
\end{align}
$$
Thus, the limit of our original function is $\cos(a)$.
A: We put $x = a + h$ so that when $x \to a+$ then $h \to 0+$. We then have
\begin{align}
L &= \lim_{x \to a+}\frac{\cos x \ln(x - a)}{\ln(e^{x} - e^{a})}\notag\\
&= \lim_{h \to 0+}\cos(a + h)\frac{\ln h}{\ln(e^{a + h} - e^{a})}\notag\\
&= \lim_{h \to 0+}\cos a \frac{\ln h}{\ln\{e^{a}(e^{h} - 1)\}}\tag{1}\\
&= \cos a\lim_{h \to 0+}\frac{\ln h}{a + \ln(e^{h} - 1)}\notag\\
&= \cos a\lim_{h \to 0+}\dfrac{\ln h}{a + \ln\left(h\cdot\dfrac{e^{h} - 1}{h}\right)}\notag\\
&= \cos a\lim_{h \to 0+}\dfrac{\ln h}{a + \ln h + \ln\left(\dfrac{e^{h} - 1}{h}\right)}\notag\\
&= \cos a\lim_{h \to 0+}\dfrac{1}{\dfrac{a}{\ln h} + 1 + \dfrac{1}{\ln h}\cdot\ln\left(\dfrac{e^{h} - 1}{h}\right)}\notag\\
&= \cos a\cdot\frac{1}{0 + 1 + 0\cdot 0}\tag{2}\\
&= \cos a\notag
\end{align}
At the end we use the fact that as $h \to 0+$, $\ln h \to -\infty$ so that $a/\ln h \to 0$ and $(e^{h} - 1)/h \to 1$ so that $\ln((e^{h} - 1)/h) \to 0$ and $1/\ln h \to 0$.
As I have pointed out elsewhere in this site, we don't need L'Hospital or series expansions for most of the limit problems.

Note: In step marked $(1)$ we assume that $\cos a\neq 0$ in order for the product rule of limits to work. And finally in step $(2)$ we see that the limit of the other factor apart from $\cos a$ is $1$ so that the condition $\cos a\neq 0$ is no longer necessary and result holds for all values of $a$. Thanks to user @LearningMath for pointing this out via comments. 
A: Since for every $x \in \mathbb{R}$ we have
$$
e^x-e^a=e^a(e^{x-a}-1)=e^a\left[x-a+\frac{(x-a)^2}{2!}+\frac{(x-a)^3}{3!}+\ldots\right]=(x-a)R(x;a),
$$
where
$$
R(x;a)=e^a\left[1+\frac{x-a}{2!}+\frac{(x-a)^2}{3!}+\ldots\right]
$$
it follows that for every $x>a$ we have
\begin{eqnarray}
f(x):&=&\frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}=\cos(x)\frac{\ln(x-a)}{\ln(x-a)+\ln R(x;a)}\\
&=&\frac{\cos(x)}{1+\frac{\ln R(x;a)}{\ln(x-a)}}
\end{eqnarray}
We have
$$
\lim_{x\to a^+}\frac{1}{\ln(x-a)}=0,\ \lim_{x\to a^+}\ln R(x;a)=\ln e^a=a,
$$
and we conclude that
$$
\lim_{x\to a+}\frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}=\frac{\cos a}{1+0\cdot a}=\cos a.
$$
