Question

Let $f\left( x \right)=\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}\cos \left( \dfrac{1}{x-1} \right)$ for $x\ne 1$, then
[a] $\underset{x\to 1-}{\mathop{\lim }}\,f\left( x \right)=0$
[b] $\underset{x\to 1-}{\mathop{\lim }}\,f\left( x \right)$ does not exist
[c] $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)=0$
[d] $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ does not exist.

Answer 1

Hint: Use the fact that for x<0 |x| = -x and for x>0 |x| = x ad at x = 0, |x| = 0. Use the property $\underset{x\to 1}{\mathop{\lim }}\,g(x)\cos \left( \dfrac{1}{x-1} \right)$ exists if and only if $\underset{x\to 1}{\mathop{\lim }}\,g\left( x \right)=0$ and
use $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)$ and $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)$.

Complete step-by-step answer:
Observe that
$\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ exists if and only if $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$ equals 0 and $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ exists if and only if $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$ equals 0. Using these properties evaluate $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$ and $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$ and hence find whether $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$ and $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ exist or not

Here $g\left( x \right)=\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$ and $f\left( x \right)=g(x)\cos \left( \dfrac{1}{x-1} \right)$.
We have LHL $=\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,g\left( x \right)$
Using $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)$, we get
LHL $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( 1-h \right)\left( 1+\left| -h \right| \right)}{\left| -h \right|}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( 1-h \right)\left( 1+h \right)}{h}$
Using $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get
LHL $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( 1-{{h}^{2}} \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,-h=0$
RHL $=\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}$
Using $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)$, we get
RHL $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( 1+h \right)\left( 1+\left| h \right| \right)}{\left| h \right|}$
Using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$, we get
RHL $=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\left( 1+2h+{{h}^{2}} \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-1-2h-{{h}^{2}}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-2h-{{h}^{2}}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\left( -2-h \right)=-2$
Since $\text{LHL}\ne \text{RHL}$we have the limit of g(x) does not exist at x = 1.
Hence the limit of f(x) does not exist at 1.
Also since LHL of g(x) = 0, we have LHL of f(x) = 0 at x=1.
Hence $\underset{x\to 1-}{\mathop{\lim }}\,f\left( x \right)=0$.
Hence option [a] is correct.
Since the RHL of g(x) is non-zero, the RHL of f(x) does not exist at x = 1. Hence option [d] is correct.
Hene options [a] and [d] are correct.

Note: [1] A limit of a function is said to be defined if and only if the Left-Hand Limit (LHL) is equal to the Right-Hand Limit(RHL),i.e. limit exists if and only if LHL = RHL.
 [2] $\underset{x\to 1}{\mathop{\lim }}\,g(x)\cos \left( \dfrac{1}{x-1} \right)$ exists if and only if $\underset{x\to 1}{\mathop{\lim }}\,g\left( x \right)=0$
This is true because if the LHL or RHL of g(x) comes out to be non-zero, then the LHL or RHL of $g(x)\cos \left( \dfrac{1}{x-1} \right)$ will fail to exist as the latter quantity will be an oscillatory quantity. Hence either LHL or RHL will fail to exist, and hence the limit of the whole function will not exist.