What is the value for lim?
I solved it by expanding \sin x as
\sin x = x - \frac {x^3} {3!} \dotsc
So \lim \limits _{x\to\infty} \frac {\sin x} x = 1 -\infty = - \infty,
but the answer is 0. Why? What I am doing wrong?
Answer
Yes , the answer is 0 .
One way to see this is by using the inequality :
\left |\frac{\sin x}{x}\right | \leq \frac{1}{x} when x>0 (this happens because |\sin x\ | \leq 1 )
When x \to \infty we have \frac{1}{x} \to 0 so the limit must be 0 .
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