SOLVED:Use the Squeeze Theorem to find each limit. limx →0[x^2(1-cos(1)/(x))]

Squeeze theorem (sandwich theorem) | Limits | Differential Calculus | Khan Academy
Squeeze theorem (sandwich theorem) | Limits | Differential Calculus | Khan Academy

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Chapter 1, Problem 49

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Use the Squeeze Theorem to find each limit.$$\lim _{x \rightarrow 0}\left[x^{2}\left(1-\cos \frac{1}{x}\right)\right]$$

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03:10

Use the Squeeze Theorem to find each limit.
$$
\lim _{x \rightarrow 0}\left[x\left(1-\cos \frac{1}{x}\right)\right]
$$

01:46

Use the Squeeze Theorem to calculate the limit.
$$\lim _{x \rightarrow \infty} \frac{1}{x+2 \cos ^{2} x}$$

01:26

Use the Squeeze Theorem to calculate the limit.
$$\lim _{x \rightarrow \infty} \frac{\cos ^{2} x}{2 x+1}$$

02:36

Find the limit and confirm your answer using the Squeeze Theorem.
$$
\lim _{x \rightarrow \infty} \frac{1-\cos x}{x^{2}}
$$

Transcript

for this problem we need to use squeeze theorem to determine the limit of X squared times one minus co sign of one over X as X approaches zero. And we begin by setting G of X equal to X squared times one minus co sign of one over X. And then from here you want to find functions F. Of X and H of X such that F of X is less than or equal to G. Of X, and G F X is less than or equal to H. Of X. To do this, we start from the fact that co sign of one over X. This is less than or equal to one, But this is greater than or equal to -1. And then if we multiply negative one to this inequality we have One greater than or equal to negative co sign of one over X. This is greater than or equal to -1. And then if we add one to this inequality we have two that’s greater than or equal to one minus go sign of one over X. This is greater than or equal to zero. Now this tells us that one minus go sign of one over X. The value of this It’s less than equal to two. And if we…

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