Answer:
See Below.
Step-by-step explanation:
We want to use the Squeeze Theorem to show that:
[tex]\displaystyle \lim_{x \to 0}\left(x^2\sin\left(\frac{2}{x}\right)\right)=0[/tex]
Recall that according to the Squeeze Theorem, if:
[tex]\displaystyle g(x)\leq f(x) \leq h(x)[/tex]
And:
[tex]\displaystyle \lim_{x\to c}g(x) =\lim_{x\to c}h(x) = L[/tex]
Then:
[tex]\displaystyle \lim_{x\to c}f(x)=L[/tex]
Recall that the value of sine is always ≥ -1 and ≤ 1. Hence:
[tex]\displaystyle -1 \leq \sin\left(\frac{2}{x}\right) \leq 1[/tex]
We can multiply both sides by x². Since this value is always positive, we do not need to change the signs. Hence:
[tex]\displaystyle -x^2\leq x^2\sin\left(\frac{2}{x}\right)\leq x^2[/tex]
Let g = -x², h = x², and f = x²sin(2 / x). We can see that:
[tex]\displaystyle \lim_{x \to 0}g(x) = \lim_{ x \to 0}h(x) = 0[/tex]
And since g(x) ≤ f(x) ≤ h(x), we can conclude using the Squeeze Theorem that:
[tex]\displaystyle \lim_{x \to 0}f(x) = \lim_{x \to 0}x^2\sin\left(\frac{2}{x}\right)=0[/tex]
