Question

$\text{Let}f\left(x\right)=\frac{1-x(1+|1-x|)}{|1-x|}\cos \left(\frac{1}{1-x}\right)$ for $x\ne 1.$ Then

• A. $\underset{x\to 1^{-}}{lim}f\left(x\right)=0$
• B. $\underset{x\to 1^{-}}{lim}f\left(x\right)$ does not exist
• C. $\underset{x\to 1^{+}}{lim}f\left(x\right)=0$
• D. $\underset{x\to 1^{+}}{lim}f\left(x\right)$ does not exist

Answer 1

Answer: (A), (D)

The correct options are
A $\underset{x\to 1^{-}}{lim}f\left(x\right)=0$
D $\underset{x\to 1^{+}}{lim}f\left(x\right)$ does not exist

$f\left(x\right)=\frac{1-x(1+|1-x|)}{|1-x|}\cos \left(\frac{1}{1-x}\right)\text{for}x\ne 1.$
$f\left(x\right)=\{\begin{array}{c}(1-x)\cos \left(\frac{1}{1-x}\right);x<1\\ -(1+x)\cos \left(\frac{1}{1-x}\right);x>1\end{array}$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}-(2+h)\cos \left(\frac{1}{-h}\right)$
Therefore RHL at $x=1$ does not exist.
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}h\cos \frac{1}{h}=0$