Solving limit without l'Hospitals rule $\lim_{n\to \infty} (\frac{\sin \frac{e}{n}}{\sin \frac{1}{n}} + \frac{\ln{\pi n}}{\ln{n}})$

Question

I have to solve this limit without using l'Hospital rule, the answer (according to Wolfram Alpha) is $e + 1$:

$$\lim_{n\to \infty} (\frac{\sin \frac{e}{n}}{\sin \frac{1}{n}} + \frac{\ln{\pi n}}{\ln{n}})$$

I know how to solve the left side of the problem:

$$\frac{\sin \frac{e}{n}}{\sin \frac{1}{n}} \rightarrow \frac{\sin \frac{e}{n}}{\frac{e}{n}} \times \frac{\frac{1}{n}}{\sin \frac{1}{n}} \times \frac{\frac{e}{n}}{\frac{1}{n}} \rightarrow e$$

But I can't find a method to solve this part of the limit:

$$\lim_{n\to \infty} \frac{\ln{\pi n}}{\ln{n}}$$

Answer 1

Hint: $\ln(ab) = \ln(a) + \ln(b)$