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Calculate $\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x$ without using l'Hopital's rule.

I attempted pulling something out to get a limit that resembles $e$, this gave me:

$$\sin^x\frac{1}{x}(1+\frac{1}{\tan\frac{1}{x}})^x=\sin^x\frac{1}{x}\bigg[(1+\frac{1}{\tan\frac{1}{x}})^{\tan\frac{1}{x}}\bigg]^{\frac{x}{\tan\frac{1}{x}}}$$

But then the first part goes to $0$ and the exponent goes to infinity, and I'm kinda stumped.

anomaly
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Nescio
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    We seem to get a lot of questions about computing various limits "without l'Hopital's rule." Seems easier to just go ahead and use l'Hopital's rule. – anomaly Jun 29 '15 at 14:24
  • @anomaly I agree, but I'm preparing for a test, and the instructions say "without l'Hopital's rule" – Nescio Jun 29 '15 at 14:29
  • I figured that; my point is that it's silly to artificially prevent students from using a simple (to both use and prove) tool that's perfectly suited to this sort of problem. – anomaly Jun 29 '15 at 14:43

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