Question

The value of f(0) so that the function $f\left( x \right)=\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}},x=0$ continuous at $x=0$ , is
(a) 1
(b) 0
(c) 2
(d) -1

Answer 1

Hint: To find value of f(0) when the function $f\left( x \right)=\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}}$ is continuous at $x=0$ , we will use the rule that a function is continuous at $x=c$ if $f\left( c \right)=\displaystyle \lim_{x \to c}f\left( x \right)$ .
We will substitute $x=0$ in this rule and find the limits. If the limit is in $\dfrac{0}{0}$ form , then we will have to use L-Hospital’s rule.

Complete step by step solution:
We have to find the value of f(0) so that the function $f\left( x \right)=\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}}$ continuous at $x=0$ . We know that a function is continuous at $x=c$ if
$f\left( c \right)=\displaystyle \lim_{x \to c}f\left( x \right)$
We are given that $f\left( x \right)=\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}}$ is continuous at $x=0$ . Then, we can write
$f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}}$
Let us apply the limit. We know that $\displaystyle \lim_{x \to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=\dfrac{\displaystyle \lim_{x \to 0}f\left( x \right)}{\displaystyle \lim_{x \to 0}g\left( x \right)}$
$\begin{align}
  & \Rightarrow f\left( 0 \right)=\dfrac{\log \left( 1+0\cdot \tan 0 \right)}{\sin 0} \\
 & \Rightarrow f\left( 0 \right)=\dfrac{\log \left( 1 \right)}{\sin 0}=\dfrac{0}{0} \\
\end{align}$
We obtained the limit as $\dfrac{0}{0}$ form which is an indeterminate form. Thus, we have to use L-Hospital’s rule. This rule states that
$\displaystyle \lim_{x \to c}\dfrac{p\left( x \right)}{q\left( x \right)}=\displaystyle \lim_{x \to c}\dfrac{p'\left( x \right)}{q'\left( x \right)}$
The given function is of the above form. Thus, we have to differentiate the numerator and denominator.
Let us consider the limit as
$\begin{align}
  & f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{\log \left( 1+{{x}^{2}}\tan x \right)}{\sin {{x}^{3}}}=\displaystyle \lim_{x \to 0}\dfrac{p\left( x \right)}{q\left( x \right)} \\
 & \Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{p'\left( x \right)}{q'\left( x \right)}...\left( i \right) \\
\end{align}$
Here, $p\left( x \right)=\log \left( 1+{{x}^{2}}\tan x \right)$ . Let us differentiate this function with respect to x.
We know that $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}$ . Also, we will have to apply chain rule.
$\begin{align}
  & \Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \dfrac{d}{dx}\left( 1+{{x}^{2}}\tan x \right) \\
 & \Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \left[ \dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}}\tan x \right) \right] \\
\end{align}$
We know that the derivative of a constant is 0.
$\begin{align}
  & \Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \left[ 0+\dfrac{d}{dx}\left( {{x}^{2}}\tan x \right) \right] \\
 & \Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \dfrac{d}{dx}\left( {{x}^{2}}\tan x \right) \\
\end{align}$
Now, let us apply the product rule.
$\Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \left[ {{x}^{2}}\dfrac{d}{dx}\left( \tan x \right)+\tan x\dfrac{d}{dx}\left( {{x}^{2}} \right) \right]$
We know that $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .
$\begin{align}
  & \Rightarrow p'\left( x \right)=\dfrac{1}{1+{{x}^{2}}\tan x}\times \left[ {{x}^{2}}{{\sec }^{2}}x+2x\tan x \right] \\
 & \Rightarrow p'\left( x \right)=\dfrac{{{x}^{2}}{{\sec }^{2}}x+2x\tan x}{1+{{x}^{2}}\tan x}...\left( ii \right) \\
\end{align}$
Now, let us consider the denominator of the given function to be $q\left( x \right)$ .
$\Rightarrow q\left( x \right)=\sin {{x}^{3}}$
Let us differentiate the above function with respect to x. For this, we will have to use chain rule. We know that $\dfrac{d}{dx}\sin x=\cos x$ .
$\Rightarrow q'\left( x \right)=\cos x\times \dfrac{d}{dx}{{x}^{3}}$
We know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ . Therefore, the above derivative becomes
$\Rightarrow q'\left( x \right)=3{{x}^{2}}\cos x...\left( iii \right)$
Now, let us substitute (ii) and (iii) in (i).
$\begin{align}
  & \Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{{{x}^{2}}{{\sec }^{2}}x+2x\tan x}{1+{{x}^{2}}\tan x}}{3{{x}^{2}}\cos x} \\
 & \Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}{{\sec }^{2}}x+2x\tan x}{\left( 1+{{x}^{2}}\tan x \right)3{{x}^{2}}\cos x} \\
\end{align}$
Let us take ${{x}^{2}}$ outside from the numerator.
$\Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{{{x}^{2}}\left( {{\sec }^{2}}x+2\cdot \dfrac{\tan x}{x} \right)}{\left( 1+{{x}^{2}}\tan x \right)3{{x}^{2}}\cos x}$
We can now cancel ${{x}^{2}}$ from the numerator and denominator.
$\Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{{{\sec }^{2}}x+2\cdot \dfrac{\tan x}{x}}{\left( 1+{{x}^{2}}\tan x \right)3\cos x}$
We can write the above limit as
$\Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\left[ \dfrac{{{\sec }^{2}}x}{\left( 1+{{x}^{2}}\tan x \right)3\cos x}+\dfrac{2\cdot \dfrac{\tan x}{x}}{\left( 1+{{x}^{2}}\tan x \right)3\cos x} \right]$
We know that $\displaystyle \lim_{x \to c}\left[ f\left( x \right)+g\left( x \right) \right]=\displaystyle \lim_{x \to c}f\left( x \right)+\displaystyle \lim_{x \to c}g\left( x \right)$ .
$\Rightarrow f\left( 0 \right)=\displaystyle \lim_{x \to 0}\dfrac{{{\sec }^{2}}x}{\left( 1+{{x}^{2}}\tan x \right)3\cos x}+\displaystyle \lim_{x \to 0}\dfrac{2\cdot \dfrac{\tan x}{x}}{\left( 1+{{x}^{2}}\tan x \right)3\cos x}$
We can write the above limit as
\[\Rightarrow f\left( 0 \right)=\dfrac{\displaystyle \lim_{x \to 0}{{\sec }^{2}}x}{\displaystyle \lim_{x \to 0}\left( 1+{{x}^{2}}\tan x \right)3\cos x}+\dfrac{2\cdot \displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}}{\displaystyle \lim_{x \to 0}\left( 1+{{x}^{2}}\tan x \right)3\cos x}\]
Let us apply the limits. We know that \[\displaystyle \lim_{x \to 0}\dfrac{\tan x}{x}=1\] .
\[\begin{align}
  & \Rightarrow f\left( 0 \right)=\dfrac{{{\sec }^{2}}0}{\left( 1+0\cdot \tan 0 \right)3\cos 0}+\dfrac{2\times 1}{\left( 1+0\cdot \tan 0 \right)3\cos 0} \\
 & \Rightarrow f\left( 0 \right)=\dfrac{1}{1\times 3\times 1}+\dfrac{2}{1\times 3\times 1} \\
 & \Rightarrow f\left( 0 \right)=\dfrac{1+2}{1\times 3\times 1}=\dfrac{3}{3}=1 \\
\end{align}\]
Therefore, the value of $f\left( 0 \right)$ is 1.

So, the correct answer is “Option A”.

Note: Students must read the values carefully as they may misunderstand $\sin {{x}^{3}}$ as ${{\sin }^{3}}x$ . They must know the derivatives of basic functions and the properties of the derivatives. They should also know the properties of limits and L'Hospital's rule. We took ${{x}^{2}}$ outside from the numerator because when we apply the limits before doing this, again the limit will be of the indeterminate form.