Question

If $f\left(a\right)=2$, $f^{\prime }\left(a\right)=1$, $g\left(a\right)=-1$ and $g^{\prime }\left(a\right)=2$, the value of
$\underset{x\to a}{lim}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}$ is

• A. $-5$
• B. $\frac{1}{5}$
• C. $5$
• D. none of these

Answer 1

The correct option is C $5$

$\underset{x\to a}{lim}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}$
$=\underset{x\to a}{lim}\frac{g\left(x\right)(f\left(a\right)-f\left(x\right))+f\left(x\right)(g\left(x\right)-g\left(a\right))}{x-a}$
$=\underset{x\to a}{lim}g\left(x\right)\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{x-a}+\underset{x\to a}{lim}f\left(x\right)\underset{x\to a}{lim}\frac{g\left(x\right)-g\left(a\right)}{x-a}$
Since $g^{\prime }\left(a\right)$ exists, g is continuous at $x=a$.
Also, $f^{\prime }\left(a\right)$ exists, so that f is continuous at $x=a.$
Hence $\underset{x\to a}{lim}g\left(x\right)=g\left(a\right)=-1$ and $\underset{x\to a}{lim}f\left(x\right)=f\left(a\right)=2$.
Thus
$\underset{x\to a}{lim}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}=-g\left(a\right)f^{\prime }\left(a\right)+f\left(a\right)g^{\prime }\left(a\right)=-(-1)1+2\times 2=5$