Find $\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}$ using L'Hospital's Rule. I am having a hard time figuring out how to find $\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)}$ using L'Hospital's Rule. I differentiate the numerator and denominator, but I'm not sure what to do with the result. This is my first time working with a limit where $x \to a^+$ instead of a known number like zero. If I just plug in $a^+$ the answer seems to become $\cos(a)$ but I need to use L'Hospital's Rule.

$$\lim_{x \to a^+} \frac{\cos(x)\ln(x-a)}{\ln(e^x-e^a)} = \lim_{x \to a^+} \frac{\frac{d}{dx}\cos(x)\ln(x-a)}{\frac{d}{dx}\ln(e^x-e^a)}$$

$$\frac{d}{dx} \ln(x - a)=\frac{x}{x - a} \to \frac{d}{dx}\cos(x)\ln(x-a) = -\sin(x)\ln(x-a)+\frac{\cos(x)}{x-a}$$

$$\frac{d}{dx}\ln(e^x-e^a) = \frac{e^x}{e^x-e^a}$$

$$\lim_{x \to a^+} \frac{\frac{d}{dx}\cos(x)\ln(x-a)}{\frac{d}{dx}\ln(e^x-e^a)} = \lim_{x \to a^+}\frac{-\sin(x)\ln(x-a)+\frac{\cos(x)}{x-a}}{\frac{e^x}{e^x-e^a}}$$

This is where I get stuck. How can I handle this?
Set $x-a=h$ in $$\dfrac{\ln(x-a)}{a+\ln(e^{x-a}-1)}$$

Then apply L'hospital rule
The cosine can be set to cos(a), while the remaining
can be calculated with the rule.
Applied twice -first we get
exp (-x) (exp x - exp a)/(x - a)
Then second time - we find exp(a-x).
Which has limit 1
Hence the final answer is:
cos(a)