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Consider the given function, \[f(x)=\left[ \begin{matrix}
   \dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{x}^{2}}-{{\left[ x \right]}^{2}}} \\
   1 \\
   \sqrt{\left\{ x \right\}\cot \left\{ x \right\}} \\
   for\text{ }x>0\text{ } \\
   for\text{ }x=0\text{ } \\
   for\text{ }x<0\text{ } \\
\end{matrix} \right.\]
where$\left[ x \right]$is the step up function and$\left\{ x \right\}$ is the fractional part function of$x$, then:
(a) $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1$
(b)$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=1$
(c) ${{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}=1$
(d) $f\text{ is continuous at }x=1$

Last updated date: 29th Feb 2024
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IVSAT 2024
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Hint: Apply limit to the given function separately at point x = 0 and x = 1 and then substitute \[\{x\}+\left[ x \right]=x\], $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1$ and $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$, simplify it further. Then check the validity of the options by using various properties of the limit.

Complete step by step answer:
  We are given the function \[f(x)=\left[ \begin{matrix}
   \dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{x}^{2}}-{{\left[ x \right]}^{2}}} \\
   1 \\
   \sqrt{\left\{ x \right\}\cot \left\{ x \right\}} \\
   for\text{ }x>0\text{ } \\
   for\text{ }x=0\text{ } \\
   for\text{ }x<0\text{ } \\
\end{matrix} \right.\]
We will apply the limit to the given function around the point \[x=0\] under various conditions and then check the continuity of the function around the point \[x=1\].
We know that \[\left\{ x \right\}\] is the function that evaluates the fractional value of \[x\] and $\left[ x \right]$ is the function that evaluates the integral value of \[x\].
For\[x>0\], we have the function\[f(x)\]such that\[f\left( x \right)=\dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{x}^{2}}-{{\left[ x \right]}^{2}}}\].
Thus, by applying the limit on the given function, we get
$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{x}^{2}}-{{\left[ x \right]}^{2}}}$
Now we will apply left hand limit using the formula, $f\left( {{0}^{+}} \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f(0+h)-f(0)}{0+h}$, we get
  & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left\{ 0+h \right\}}{{{\left( 0+h \right)}^{2}}-{{\left[ 0+h \right]}^{2}}} \\
 & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}h}{{{h}^{2}}} \\
As, we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\tan x}{x}=1$, so, we get \[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}x}{{{x}^{2}}}=1\] as well.
Thus, we get \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{\left\{ x \right\}}^{2}}}=1\].
Hence, we have the value of limit as \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{{{\tan }^{2}}\left\{ x \right\}}{{{x}^{2}}-{{\left[ x \right]}^{2}}}=1\].
Now, we will consider the case \[x<0\]. For \[x<0\], we have the function \[f(x)\] such that \[f\left( x \right)=\sqrt{\left\{ x \right\}\cot \left\{ x \right\}}\].
Applying the limit on the given function, we get
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\{ x \right\}\cot \left\{ x \right\}}$
Further simplifying the limit, we have
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\{ x \right\}\dfrac{\cos \left\{ x \right\}}{\sin \left\{ x \right\}}}$ as we know that \[\cot x=\dfrac{\cos x}{\sin x}\].
As we know that $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1$, we have\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\{ x \right\}\dfrac{\cos \left\{ x \right\}}{\sin \left\{ x \right\}}}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\cos \left\{ x \right\}}\]
Thus, we have
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\cos \left\{ x \right\}}=\sqrt{\cos 0}=1\]
Hence, we have
\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\{ x \right\}\cot \left\{ x \right\}}=1\]
Now, we need to evaluate the value of
${{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}$
As \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\sqrt{\left\{ x \right\}\cot \left\{ x \right\}}=1\], we have
\[{{\cot }^{-1}}{{\left( \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x) \right)}^{2}}={{\cot }^{-1}}{{\left( 1 \right)}^{2}}={{\cot }^{-1}}1=\dfrac{\pi }{4}\]
Now, we will check the continuity of \[f\] at point $x=1$ as for \[x=1\], we have
\[f\left( {{x}^{-}} \right)=f\left( {{x}^{+}} \right)\].
Thus, the function\[f\]is continuous at $x=1$

So, the correct answers are “Option A, B and D”.

Note: It’s necessary to evaluate both left- and right-hand side of the limit around a point. Otherwise, we won’t get a correct answer if we apply only one side of the limit.
Students sometimes substitute \[\{x\}+\left[ x \right]=x\Rightarrow \left[ x \right]=x-\{x\}\], in this way the process will get lengthy and chances of getting the wrong solution is there.
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