# Math Help - [SOLVED] finding a limit

1. ## [SOLVED] finding a limit

$\lim_{x \to 0} (e^x-x-1) \cos^2 \frac{2}{x} + x + 1$

I know that I need to use the squeeze theorem, and I have an explanation of how to solve the problem, but I don't understand one of the steps, how does:

$0 \leq \cos^2 \frac{2}{x} \leq 1 \Rightarrow$
$x+1 \leq (e^x-x-1)\cos^2 \frac{2}{x}+x+1 \leq e^x$ ?

2. The cosine varies between -1 and 1, so the square, being non-negative, varies between 0 and 1. Now do the plug-n-chug:

If you take the lower limit, then (exponential stuff)*(cosine stuff) is bounded below by (exponential stuff)*(zero), or zero.

If you take the upper limit, then (exponential stuff)*cosine stuff) is bounded above by (exponential stuff)*(one), or one.

If you're dealing with the lower limit, the expression simplifies as 0 + x + 1 = x + 1.

If you're dealing with the upper limit, the expression simplified as (e^x - x - 1) + x + 1 = e^x + x - x + 1 - 1 = e^x.