Question

If $f\left(a\right)=2$, $f^{{}^{\prime }}\left(a\right)=1$, $g\left(a\right)=1$, $g^{{}^{\prime }}\left(a\right)=-2$, then $\stackrel{lim}{x\to 0}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}$, is

Answer 1

$f\left(a\right)=2$, $f^{{}^{\prime }}\left(a\right)=1$, $g\left(a\right)=1$, $g^{{}^{\prime }}\left(a\right)=-2$,
$L=\stackrel{lim}{x\to a}\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}=\frac{(-1)\left(2\right)-(-1)\left(2\right)}{x-a}=\frac{0}{0}L=\stackrel{lim}{x\to a}\frac{g^{{}^{\prime }}\left(x\right)f\left(a\right)-g\left(a\right)f^{{}^{\prime }}\left(x\right)}{1}=(-2)\left(2\right)-(-1)\left(1\right)=-4+1=-3$