Question

Examine the continuity of the function $f\left(x\right)=\{\begin{array}{cc}|x|\cos \frac{1}{x}& ,\text{if}x\ne 0\\ 0& ,\text{if}x=0\end{array}$ at $x=0$

Answer 1

Given:

$f\left(x\right)=\{\begin{array}{c}x\cos \frac{1}{x};x\ne 0\\ 0;x=0\end{array}$

To show that $f\left(x\right)$ is continuous at $x=0$

or ${lim}_{x\to 0}f\left(x\right)=f\left(0\right)=0$

Now we have to evaluate

${lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}x\cos \frac{1}{x}$

As we know that for all $x\in R-\left\{0\right\},R$ is the set of real numbers.

$-1\le \cos \frac{1}{x}\le 1$

$\Rightarrow -x\le x\cos \frac{1}{x}\le x$ for $x>0$ ..........$\left(1\right)$

and $x\le x\cos \frac{1}{x}\le -x$ for $x<0$ ..........$\left(2\right)$

Now, ${lim}_{x\to 0}(-x)=0$

and ${lim}_{x\to 0}x=0$

From $\left(1\right)$ and $\left(2\right)$, we have

${lim}_{x\to 0}x\cos \frac{1}{x}=0$ by sandwich theorem

Thus,${lim}_{x\to 0}f\left(x\right)=f\left(0\right)=0$ is satisfied.

Hence $f\left(x\right)$ is continuous at $x=0$

Hence proved.