Question

lf $f\left(a\right)=2,f^{\prime }\left(a\right)=1,g\left(a\right)=-1,g^{\prime }\left(a\right)=2$, then $\underset{x\to a}{lim}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}$=

• A. $\frac{1}{5}$
• B. $5$
• C. $-\frac{1}{5}$
• D. $-5$

Answer 1

Answer: (B)

$5$

Given, $f\left(a\right)=2,f^{\prime }\left(a\right)=1$
$g\left(a\right)=-1,g^{\prime }\left(a\right)=2$
$L={lim}_{x\to a}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}$
Since it is of $\frac{0}{0}$ form, using L'Hospital's rule, we get
$L={lim}_{x\to a}f\left(a\right)g^{\prime }\left(x\right)-g\left(a\right)f^{\prime }\left(x\right)=g^{\prime }\left(a\right)f\left(a\right)-g\left(a\right)f^{\prime }\left(a\right)=\left(2\right)\left(2\right)-(-1)\left(1\right)=5$