MHB Unraveling the Complexity of a Limit Problem Involving ln(x)

[Dethrone]

I'm having trouble with this limit. It gets complicated really quickly when I apply l'hopital's. Any hints?

$$\lim_{{x}\to{a^+}}\frac{\cos\left({x}\right)\ln\left({x-a}\right)}{\ln\left({e^x-e^a}\right)}$$

 

[ineedhelpnow]

Is the problem out of a book?

 

[Dethrone]

All the problems I post are from past Calc I exam papers. This question is worth 3 marks out of 100. Don't see why it would be so difficult (Headbang)

 

[MarkFL]

Hint:

$$\lim_{x\to a}\left[f(x)\cdot g(x)\right]=\lim_{x\to a}f(x)\cdot\lim_{x\to a}g(x)$$

edit: Never mind, the post below gives the complete solution.

 

[chisigma]

I'm having trouble with this limit. It gets complicated really quickly when I apply l'hopital's. Any hints?

$$\lim_{{x}\to{a^+}}\frac{\cos\left({x}\right)\ln\left({x-a}\right)}{\ln\left({e^x-e^a}\right)}$$

The term cos x is not critical and You can look at the limit of...

$\displaystyle f(x) = \frac{\ln (x-a)}{\ln (e^{x} - e^{a})}\ (1)$

Now applying two times de l'Hopital You obtain...

$\displaystyle \lim_{x \rightarrow a+} \frac{\ln (x-a)}{\ln (e^{x} - e^{a})} = \lim_{x \rightarrow a+} e^{-x} \frac{e^{x} - e^{a}}{x - a} = \lim_{x \rightarrow a+} e^{-x}\ e^{x} = 1\ (2)$

... so that Your limit is [simply] cos a...

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

Let me see...I can't believe I didn't see that...

$$=\cos\left({a}\right)\lim_{{x}\to{a^+}}\frac{\ln\left({x-a}\right)}{\ln\left({e^x-e^a}\right)}$$
$$=\frac{\cos\left({a}\right)}{e^a}\lim_{{x}\to{a^+}}\frac{e^x-e^a}{x-a}$$
Applying my favorite trick :D
$$=\frac{\cos\left({a}\right)}{e^a}\d{}{x}e^x|_{x=a}$$
$$=\cos\left({a}\right)$$

A lucky coincidence. If you wrongly assume that $\ln\left({x-a}\right)=\ln\left({x}\right)-\ln\left({a}\right)=\ln\left({\frac{x}{a}}\right)$, you get the same answer. (Dull)