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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.

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  • $\begingroup$ hint: use squeeze theorem to prove it $\endgroup$ – K.K.McDonald Apr 17 '16 at 1:23
  • $\begingroup$ Hint: Notice that $-\sqrt{1+t^2} \leq \cos(n)+t\sin(n) \leq \sqrt{1+t^2}$, for every $t>0$, and for every $n$. The Squeeze Theorem would finish it. $\endgroup$ – Nicholas Stull Apr 17 '16 at 1:28
  • $\begingroup$ Oh that makes a lot of sense; thanks a ton for all the help, everyone! $\endgroup$ – KR136 Apr 17 '16 at 1:30
  • $\begingroup$ By the way, while the assumption $t>0$ is absolutely essential to proving the limit, it is not needed for the bounds I gave. And if you used the bounds $-1-|t|\leq \cos(n)+t\sin(n)\leq 1+|t|$, this is also true for $t\in\mathbb{R}$ (and is a simple modification of the bounds used in the answer below) $\endgroup$ – Nicholas Stull Apr 17 '16 at 1:32
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Note that $-1-t\leq\cos(n)+t\sin(n)\leq 1+t$

So we can use the squeeze theorem:

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

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