Question

$\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$, given that $f^{\prime }\left(2\right)=6$ and $f^{\prime }\left(1\right)=4$.

• A. Does not exist
• B. Is equal to $-3/4$
• C. Is equal to $3$
• D. Is equal to $3/2$

Answer 1

The correct option is C Is equal to $3$

Given $\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$
Applying L'hopital's Rule
$\underset{h\to 0}{lim}\frac{f^{\prime }(2h+2+h^2)(2+2h)-0}{f^{\prime }(h-h^2+1)(1-2h)-0}$
$=\frac{f^{\prime }\left(2\right)(2+2(0\left)\right)}{f^{\prime }(+1)(1-2(0\left)\right)}$
$=\frac{6\left(2\right)}{4}$ from given values
$=3$
Hence, the correct answer is $3$