Question

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

• A. $3/2$
• B. $3$
• C. $5/2$
• D. $-3$

Answer 1

The correct option is B $3$

$\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$
$=\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{2h+2+h^2-2}\times \frac{h(2+h)}{h(1-h)}\times \frac{(h-h^2+1)-1}{f(h-h^2+1)-f\left(1\right)}$

$=f^{\prime }\left(2\right)\times \underset{h\to 0}{lim}\frac{2+h}{1-h}\times \frac{1}{f^{\prime }\left(1\right)}=6\times 2\times \frac{1}{4}=3$
Note that L Hospital's Rule is not applicable in this case.