Question

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

Answer 1

Hint: First, before proceeding for this, we must solve it according to the given options to get the correct answer. Then, starting with the condition $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, we get $1-x<0$, we get the value of this limit. Then, proceeding with the condition $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, we get $1-x > 0$,we get the value of this limit. Then we get the answers for both the limit conditions.

Complete step-by-step answer:
In this question, we are supposed to find the condition for f(x) when $f\left( x \right)=\dfrac{1-x\left( 1+\left| 1-x \right| \right)}{\left| 1-x \right|}\cos \left( \dfrac{1}{1-x} \right)$.
So, before proceeding for this, we must solve it according to the given options to get the correct answer.
Then, starting with the condition $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$, we get:
$\begin{align}
  & x>1 \\
 & \Rightarrow 1-x<0 \\
\end{align}$
So, the above condition gives the value of $\left| 1-x \right|$, we get the value as $-\left( 1-x \right)$, then by substituting this value in the given limit function as:
$\begin{align}
  & f\left( x \right)=\dfrac{1-x\left( 1-\left( 1-x \right) \right)}{x-1}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{1-x\left( 1-1+x \right)}{x-1}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{1-x\left( x \right)}{x-1}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{1-{{x}^{2}}}{x-1}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{\left( 1-x \right)\left( 1+x \right)}{x-1}\cos \left( \dfrac{1}{1-x} \right) \\
\end{align}$
Then, after substituting the limit as 1 stated in the option, we get:
$\begin{align}
  & f\left( 1 \right)=\dfrac{\left( 1-1 \right)\left( 1+1 \right)}{1-1}\cos \left( \dfrac{1}{1-1} \right) \\
 & \Rightarrow f\left( 1 \right)=2\cos \left( \dfrac{1}{0} \right) \\
\end{align}$
Here, we know that $\cos \left( \dfrac{1}{0} \right)$value does not exist, so the value of $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$does not exist.
Similarly, by proceeding with the condition $\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f\left( x \right)$, we get:
$\begin{align}
  & x<1 \\
 & \Rightarrow 1-x>0 \\
\end{align}$
So, the above condition gives the value of $\left| 1-x \right|$, we get the value as $\left( 1-x \right)$, then by substituting this value in the given limit function as:
\[\begin{align}
  & f\left( x \right)=\dfrac{1-x\left( 1+\left( 1-x \right) \right)}{1-x}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{1-x\left( 1+1-x \right)}{1-x}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{1-2x+{{x}^{2}}}{1-x}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=\dfrac{{{\left( x-1 \right)}^{2}}}{1-x}\cos \left( \dfrac{1}{1-x} \right) \\
 & \Rightarrow f\left( x \right)=-\left( x-1 \right)\cos \left( \dfrac{1}{1-x} \right) \\
\end{align}\]
Then, after substituting the limit as 1 stated in the option, we get:
$\begin{align}
  & f\left( 1 \right)=-\left( 1-1 \right)\cos \left( \dfrac{1}{1-1} \right) \\
 & \Rightarrow f\left( 1 \right)=0\cos \left( \dfrac{1}{0} \right) \\
 & \Rightarrow f\left( 1 \right)=0 \\
\end{align}$
Here, the value of $\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ is zero.
Hence, option (b) and (c) are correct.

Note: Now, to solve these types of questions we need to know some of the basic conversions beforehand so that we can easily proceed in these types of questions. So, the basic conversions are:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. Moreover, in this type of question, we don’t check after getting a single option as correct but sometimes we get more than one option as correct.