(c) use the squeeze theorem to find $lim_{x\rightarrow0}xsinleft(\frac{10}{x}\right)$. $lim_{x\rightarrow0}xs…

Explanation:

Step1: Recall the range of sine function

We know that $- 1\leqslant\sin\left(\frac{10}{x}\right)\leqslant1$ for all $x\neq0$.

Step2: Multiply the inequality by $x$

When $x>0$, we have $-x\leqslant x\sin\left(\frac{10}{x}\right)\leqslant x$. When $x < 0$, we have $x\leqslant x\sin\left(\frac{10}{x}\right)\leqslant - x$.

Step3: Find the limits of the bounding - functions

We know that $\lim_{x\rightarrow0}(-x)=0$ and $\lim_{x\rightarrow0}(x) = 0$.

Step4: Apply the squeeze theorem

By the squeeze theorem, since $\lim_{x\rightarrow0}(-x)=\lim_{x\rightarrow0}(x)=0$ and $-|x|\leqslant x\sin\left(\frac{10}{x}\right)\leqslant|x|$, then $\lim_{x\rightarrow0}x\sin\left(\frac{10}{x}\right)=0$.

Answer:

$0$