Question

suppose we wish to evaluate $lim_{x \to 0}x^{10}sin(\frac{1}{x}) - 6$. since we know that $- 1leqsin(\theta)leq1$ for $\thetainmathbb{r}$, we must have $-1leqsin(\frac{1}{x})leq1$ for $x
eq0$. from this, we must have $-1leq x^{10}sin(\frac{1}{x})leq1$ for $x
eq0$. $-1leq x^{10}sin(\frac{1}{x}) - 6leq1$ for $x
eq0$.

Explanation:

Step1: Multiply inequality by $x^{10}$

Since $- 1\leqslant\sin(\frac{1}{x})\leqslant1$ for $x
eq0$ and $x^{10}\geqslant0$ for all real - valued $x$, when we multiply the inequality by $x^{10}$, we get $-x^{10}\leqslant x^{10}\sin(\frac{1}{x})\leqslant x^{10}$.

Step2: Subtract 6 from the new inequality

Subtracting 6 from each part of the inequality $-x^{10}\leqslant x^{10}\sin(\frac{1}{x})\leqslant x^{10}$, we have $-x^{10}-6\leqslant x^{10}\sin(\frac{1}{x}) - 6\leqslant x^{10}-6$.

Step3: Find the limit as $x\to0$

We know that $\lim_{x\to0}(-x^{10}-6)=-6$ and $\lim_{x\to0}(x^{10}-6)=-6$. By the Squeeze Theorem, if $f(x)\leqslant g(x)\leqslant h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $x = a$) and $\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L$, then $\lim_{x\to a}g(x)=L$. So, $\lim_{x\to0}(x^{10}\sin(\frac{1}{x})-6)=-6$.

Answer:

$-6$