Question

If $f\left(a\right)=2,f\text{'}\left(a\right)=1,g\left(a\right)=3,g\text{'}\left(a\right)=-1$then $\underset{x\to a}{\lim }\frac{[f(a)g(x)-f(x)g(a\left)\right]}{(x-a)}=$

• A. $6$
• B. $1$
• C. $-1$
• D. $-5$

Answer 1

The correct option is D

$-5$

Explanation for the correct option:

Step 1. Find the value of $\underset{x\to a}{\lim }\frac{[f(a)g(x)-f(x)g(a\left)\right]}{(x-a)}$

Given data: $f\left(a\right)=2,f\text{'}\left(a\right)=1,g\left(a\right)=3,g\text{'}\left(a\right)=-1$

$\Rightarrow$$\begin{array}{rcl}\underset{x\to a}{\lim }\frac{[f(a)g(x)-f(x)g(a\left)\right]}{(x-a)}& =& \frac{f\left(a\right)g\left(a\right)-f\left(a\right)g\left(a\right)}{\left(a-a\right)}\\ & =& \frac{0}{0}\end{array}$

Step 2: Apply L' hospital rule:

$\underset{\mathrm{x}\to \mathrm{a}}{\therefore lim}\frac{\mathrm{f}\left(\mathrm{a}\right)\mathrm{g}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{a}\right)}{\mathrm{x}-\mathrm{a}}$$=\underset{x\to a}{lim}\frac{f\left(a\right)g\text{'}\left(x\right)-f\text{'}\left(x\right)g\left(a\right)}{1-0}$ $\left[\mathbf{\because }\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\right]$

$=\underset{x\to a}{lim}f\left(a\right)g^{\prime }\left(a\right)-f^{\prime }\left(a\right)g\left(a\right)$

$$\begin{gathered} =2(-1)-\left(1\right)\left(3\right) \\ =-5 \end{gathered}$$

Hence, option (D) is correct option.