Question

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\\}} \\\ \end{matrix}\begin{matrix} 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) $f\text{ is continuous at }x=1$

Answer 1

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\\}} \\\ \end{matrix}\begin{matrix} 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 thatf\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
\begin{aligned} & \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}}} \\\ \end{aligned}
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.