calculus - Proof that the following limit equals $0$

I have the following limit:

$$\lim_{n\to\infty} \left[\frac{\cos(n)+t\sin(n)}{e^{tn}}\right]$$

I wish to show that the limit equals $0$ rigorously. I have a good sense that it equals zero by taking the Taylor series expansion of the top, and comparing with that of the bottom. (The magnitude of the terms are the same, but the numerator has alternating signs, whereas the denominator is strictly positive. I should say; assume that $t>0$.)

I'm not sure however, whether that approach can be made rigorous. Other than that I am at a loss to prove it by a theorem or definition.

Answer

Note that $-1-t\leq\cos(n)+t\sin(n)\leq 1+t$

So we can use the squeeze theorem:

$$0=\lim_{n\to\infty} \frac{-1-t}{e^{tn}}\leq \lim_{n\to\infty} \frac{\cos(n)+t\sin(n)}{e^{tn}}\leq \lim_{n\to\infty} \frac{1+t}{e^{tn}}=0$$