Question

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

• A. $3$
• B. $-\frac{3}{2}$
• C. $\frac{3}{2}$
• D. Does not exist

Answer 1

The correct option is A $3$

$\underset{h\to 0}{lim}\frac{f(2h+2+h^2)-f\left(2\right)}{f(h-h^2+1)-f\left(1\right)}$ which is in $\frac{0}{0}$ form
$=\underset{h\to 0}{lim}\frac{\left\{f^{\prime }(2h+2+h^2)\right\}(2+2h)-0}{\left\{f^{\prime }(h-h^2+1)\right\}(1-2h)-0}$ ........ (using L'Hospital's rule)

$=\underset{h\to 0}{lim}\frac{\left\{f^{\prime }(2h+2+h^2)\right\}(2+2h)}{\left\{f^{\prime }(h-h^2+1)\right\}(1-2h)}$
$=\frac{f^{\prime }\left(2\right)\cdot 2}{f^{\prime }\left(1\right)\cdot 1}=\frac{6\times 2}{4\times 1}$
$=\frac{12}{4}=3$

Hence, option A is correct.