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.

 

[math771]

To find the limit of cos(x), we can use the following inequalities:

1. -1 ≤ cos(x) ≤ 1
2. |cos(x)| ≤ 1

To find the limit as x approaches a certain value, we can use the Squeeze Theorem. This states that if f(x) ≤ g(x) ≤ h(x) for all x near a, and lim f(x) = lim h(x) = L, then lim g(x) = L.

So, for cos(x), we can use the inequality -1 ≤ cos(x) ≤ 1 as our f(x) and h(x) functions, and find the limit of those functions to be L = 1. Then, by the Squeeze Theorem, we can conclude that lim cos(x) = 1.

For part (b), we can use the same approach. We can use the inequality |cos(x)| ≤ 1 as our g(x) function, and find the limit of that function to be L = 1. Then, by the Squeeze Theorem, we can conclude that lim cos(x) = 1 as well.

I hope this helps! Let me know if you have any other questions.