Question

How do I find this limit?

Answer 1

Used the bounded-ness of cosine: \(-1\le \cos x\le1\). As \(x\to\infty\), \(\cos x\) will oscillate between these two values, so you're basically dealing with a constant in the numerator.

Answer 2

i'd look at it as:

\[\lim_{x->\infty}\left( \frac{ 1 }{ x^2 } - \frac{ cosx }{ x^2 } \right) \]

first term is zero

\[-1 \le cosx \le 1\]
\[\frac{ -1 }{ x^2 } \le \frac{ \cos }{ x^2 } \le \frac{ 1 }{ x^2 }\]

taking the limit of ± 1/x^2 tells us by squeeze theorem that the limit of cosx/x^2 is also 0