Question

\[\lim_{x \rightarrow \infty} (sinx)/ (e^x)\] I know you could argue since Sin(x) oscillates between [-1,1], then the answer is just going to be 0. But is there a way to prove it Algebraically?

Answer 1

Well, the limit of a product is the product of its limits, so\[\lim_{x \rightarrow \infty}\frac{\sin x}{e^{x}} = \left(\lim_{x \rightarrow \infty}\sin x\right)\left(\lim_{x \rightarrow \infty}\frac{1}{e^{x}} \right)\] The limit of sin x is not 0 or infinity, so it doesn't create an indeterminate form when multiplied with the limit of 1/e^x which is 0. So you would get some non-infinity-non-zero number times zero, which would be zero. I guess.

Answer 2

ooh Thank you, i was curious about this problem.