Question

If $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ is 2, where $f\left( x \right)=\dfrac{ax{{e}^{x}}-b\log \left( 1+x \right)+cx{{e}^{-x}}}{{{x}^{2}}\sin x}$ and a, b, c are real numbers. On the basis of above information answer the following questions. The value of $\underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)$ equals
(a) 0
(b) 1
(c) 3
(d) 4

Answer 1

To solve this question, we need to first find the values of the variables a, b and c. We will do this by solving the limit with the help of L’Hopital rule. In this method, we differentiate the numerator and denominator with respect to the variable of limit. This process is repeated until the function is no longer in indeterminate form. Once we are able to find the values of a, b and c, we will find the value of $\underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)$.

Complete step-by-step answer :
It is given to us that $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$ is 2 and $f\left( x \right)=\dfrac{ax{{e}^{x}}-b\log \left( 1+x \right)+cx{{e}^{-x}}}{{{x}^{2}}\sin x}$.
Now, we will solve the limit with help of L’Hopitals’ rule.
We will differentiate numerator and denominator with respect to x.
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a\left( {{e}^{x}}+x{{e}^{x}} \right)-\dfrac{b}{1+x}+c\left( {{e}^{-x}}-x{{e}^{-x}} \right)}{2x\sin x+{{x}^{2}}\cos x}$
If we substitute x = 0, we will get the function as follows:
$\Rightarrow \dfrac{a\left( 1+0 \right)-\dfrac{b}{1+0}+c\left( 1-0 \right)}{0}$
This is indeterminate form as denominator is 0. Thus, we can say that $a-b+c=0......\left( 1 \right)$
We will continue with the L’Hopitals’ rule and differentiate the numerator and denominator again.
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a\left( {{e}^{x}}+{{e}^{x}}+x{{e}^{x}} \right)+\dfrac{b}{{{\left( 1+x \right)}^{2}}}+c\left( -{{e}^{-x}}+x{{e}^{-x}}-{{e}^{-x}} \right)}{2\sin x+2x\cos x+2x\cos x-{{x}^{2}}\sin x}$
We will substitute x = 0.
$\begin{aligned} & \Rightarrow \dfrac{a\left( 2+0 \right)+\dfrac{b}{{{\left( 1+0 \right)}^{2}}}+c\left( -1-1 \right)}{0} \\\ & \Rightarrow \dfrac{2a+b-2c}{0} \\\ \end{aligned}$
Thus, with the same logic as above, $2a+b-2c=0......\left( 2 \right)$
We shall continue with L’Hopital’s rule.
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{a\left[ {{e}^{x}}\left( 2+x \right)+{{e}^{x}} \right]-\dfrac{2b}{{{\left( 1+x \right)}^{3}}}+c\left[ -{{e}^{-x}}\left( x-2 \right)+{{e}^{-x}} \right]}{2\cos x-2x\sin x+2\cos x-2x\sin x+2\cos x-2x\sin x-{{x}^{2}}\cos x}$
We will substitute x = 0
$\begin{aligned} & \Rightarrow \dfrac{a\left( 2+1 \right)-2b+c\left( 2+1 \right)}{6} \\\ & \Rightarrow \dfrac{3a-2b+3c}{6} \\\ \end{aligned}$
Since, the denominator is not 0, we have solved the limit.
$\begin{aligned} & \Rightarrow \dfrac{3a-2b+3c}{6}=2 \\\ & \Rightarrow 3a-2b+3c=12......\left( 3 \right) \\\ \end{aligned}$
Now, we have (1), (2) and (3) as three equations in the form of a, b and c. We will solve these equations to find the value of a, b and c.
From (1), a = b – c. Substitute this in equation (2).
$\Rightarrow 2\left( bc \right)+b2c=0$
$\begin{aligned} & \Rightarrow 3b-4c=0 \\\ & \Rightarrow b=\dfrac{4}{3}c \\\ \end{aligned}$
Substitute this in equation (3).

& \Rightarrow 3\left( \dfrac{4}{3}c-c \right)-\dfrac{8}{3}c+3c=12 \\\ & \Rightarrow 4c-3c-\dfrac{2}{3}c=12 \\\ & \Rightarrow c=9 \\\ & \Rightarrow b=\dfrac{4}{3}\left( 9 \right)=12 \\\ & \Rightarrow a=12-9=3 \\\ \end{aligned}$$ Therefore, $f\left( x \right)=\dfrac{3x{{e}^{x}}-12\log \left( 1+x \right)+9x{{e}^{-x}}}{{{x}^{2}}\sin x}$ Now, we will multiple both sides with $x\sin x$. $\Rightarrow x\sin xf\left( x \right)=\dfrac{3x{{e}^{x}}-12\log \left( 1+x \right)+9x{{e}^{-x}}}{x}$ Now, we apply limits x tends to 0 on both sides. $\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{3x{{e}^{x}}-12\log \left( 1+x \right)+9x{{e}^{-x}}}{x}$ We will solve this limit with L’Hopitals’ rule. $\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{3\left( {{e}^{x}}+x{{e}^{x}} \right)-\dfrac{12}{1+x}+9\left( {{e}^{-x}}-x{{e}^{-x}} \right)}{1}$ Substitute x = 0 in the expression on the RHS. $\begin{aligned} & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)=\dfrac{3\left( 1+0 \right)-\dfrac{12}{1+0}+9\left( 1-0 \right)}{1} \\\ & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)=3-12+9 \\\ & \Rightarrow \underset{x\to 0}{\mathop{\lim }}\,x\sin xf\left( x \right)=0 \\\ \end{aligned}$ **Hence, option (a) is the correct option.** **Note** : We have solved linear equations in three variables with substitution method. They can be also solved using matrix form. L’Hopitals’ rule can be used when we have the indeterminate form of $\dfrac{0}{0}$.