MHB Find Limit of cos(x) with Inequalities | Part (b) Help

[Joe20]

Need advice on how to find lim of cos(x) using the inequalities provided.

Also part (b) for help.

Thanks.

 

[HOI]

I presume you have already shown that \lim_{x\to 0} sin(x)= 0. So focus on 0\le 1- cos(x)\le \frac{sin^2(x)}{1+ cos(x)}. Assume, for the moment, that \lim_{x\to 0} cos(x) exists and is equal to "A". Then, taking the limit of each part as x goes to 0, we have 0\le 1- A\le \frac{0}{1+ A} so 0\le 1- A\le 0. What is "A"?

For (b), if you let u= x- a then the expression becomes "sin(x- a)= sin(u)cos(a)- sin(a)cos(u)" and taking the limit as x goes to a is the same as taking the limit as u goes to 0.