Question

Use the Squeeze Theorem ​

Answer 1

Answer:

See Below.

Step-by-step explanation:

We want to use the Squeeze Theorem to show that:

$\displaystyle \lim_{x \to 0}\left(x^2\sin\left(\frac{2}{x}\right)\right)=0$

Recall that according to the Squeeze Theorem, if:

$\displaystyle g(x)\leq f(x) \leq h(x)$

And:

$\displaystyle \lim_{x\to c}g(x) =\lim_{x\to c}h(x) = L$

Then:

$\displaystyle \lim_{x\to c}f(x)=L$

Recall that the value of sine is always ≥ -1 and ≤ 1. Hence:

$\displaystyle -1 \leq \sin\left(\frac{2}{x}\right) \leq 1$

We can multiply both sides by x². Since this value is always positive, we do not need to change the signs. Hence:

$\displaystyle -x^2\leq x^2\sin\left(\frac{2}{x}\right)\leq x^2$

Let g = -x², h = x², and f = x²sin(2 / x). We can see that:

$\displaystyle \lim_{x \to 0}g(x) = \lim_{ x \to 0}h(x) = 0$

And since g(x) ≤ f(x) ≤ h(x), we can conclude using the Squeeze Theorem that:

$\displaystyle \lim_{x \to 0}f(x) = \lim_{x \to 0}x^2\sin\left(\frac{2}{x}\right)=0$