Question

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

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

Answer 1

Answer: (A), (C)

$\underset{x\to 1^{-}}{lim}f\left(x\right)=0$
$\underset{x\to 1^{+}}{lim}f\left(x\right)$ does not exist

$f\left(x\right)=\frac{1-x\left(1+1-x\right)}{1-x}\cos \left(\frac{1}{1-x}\right)$ for $x<1$
$=\frac{1-x\left(2-x\right)}{1-x}\cos \left(\frac{1}{1-x}\right)$
$=\frac{1-2x+x^2}{1-x}\cos \left(\frac{1}{1-x}\right)$
and $f\left(x\right)=\frac{1-x\left(1+x-1\right)}{x-1}\cos \left(\frac{1}{1-x}\right)$ for $x>1$
$=\frac{1-x^2}{x-1}\cos \left(\frac{1}{1-x}\right)$
Now $\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}\frac{1-x^2}{x-1}\cos \left(\frac{1}{1-x}\right)$
$=\underset{x\to 1}{lim}\frac{-\left(x-1\right)\left(x+1\right)}{x-1}\cos \left(\frac{1}{1-x}\right)$
$=\underset{x\to 1}{lim}-\left(x+1\right)\cos \left(\frac{1}{1-x}\right)$
$=-2\times \underset{x\to 1}{lim}\cos \left(\frac{1}{1-x}\right)$
$=$does not esist
similarly, $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\frac{1-2x+x^2}{1-x}\cos \left(\frac{1}{1-x}\right)$
$=\underset{x\to 1}{lim}\frac{{\left(x-1\right)}^2}{-(x-1)}\cos \left(\frac{1}{1-x}\right)$
$=\underset{x\to 1}{lim}-\left(x-1\right)\cos \left(\frac{1}{1-x}\right)$
$=0\times \underset{x\to 1}{lim}\cos \left(\frac{1}{1-x}\right)$
$=0$
So, $\underset{x\to 1^{+}}{lim}f\left(x\right)$ does not exists and $\underset{x\to 1^{-}}{lim}f\left(x\right)$ $=0$