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This question already has an answer here:

Prove $\lim_{a \to \infty} \frac{sina}{a} = 0$

Attempt: $$$$

$|\frac{sina}{a} - 0| \leq |\frac{1}{a}|$ so $\frac{1}{a} < \epsilon \implies a > \frac{1}{\epsilon} $

so

$\forall \epsilon > 0 \exists N=\frac{1}{\epsilon} : \forall a> N \implies |\frac{sina}{a} - 0| \leq |\frac{1}{a}| = \frac{1}{a} < \epsilon $ which means $\lim_{a \to \infty} \frac{sina}{a} = 0$

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marked as duplicate by hardmath, k170, Daniel Fischer Nov 2 '15 at 18:57

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    $\begingroup$ Note that $1/\epsilon$ may not be an integer (not sure if you mean to implicitly assume so by assigning it to a variable called $N$). Also, $|1/x|$ seems out of place in the last chain of inequalities. (What is $x$?) $\endgroup$ – Bungo Nov 2 '15 at 18:34
  • $\begingroup$ note that $|\sin(x)|\le 1$ $\endgroup$ – Dr. Sonnhard Graubner Nov 2 '15 at 18:41