Lim([cos x ln(x-a)]/ln(e^x-e^a)) as x approaches a from the right?

Question

I need help working this problem using L'Hopital's rule. I need to find the limit as x approaches a+ for cosx ln(x-a) divided by ln(e^x-e^a).

Answer 1

This can be done without L'Hopital, but It's possible with it.

Since cos is continuous,it's limit at a+ is cosa. Hence we only need to determine lim (x → a+) ln(x-a)]/ln(e^x-e^a)). This as limit of the ∞/∞ form, so that L'Hopital is indeed applicable. Taking the derivatives of the numerator and of the denominator, we get

lim (x → a+) [1/(x -a)]/[e^x/(e^x - e^a)] = lim (x → a+) (e^x - e^a)/[e^x(x-a)] = lim (x → a+) (e^x - e^a)/(x -a) e^(-x) = lim (x → a+) (e^x - e^a)/(x -a) * lim (x → a+) e^(-x) = e^a * e^(-a) = 1. Since this limit exists, it follows that

So,   lim(x → a+)  ([cos x ln(x-a)]/ln(e^x-e^a)) = cosa * 1 = cosa