Question

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

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

Answer 1

The correct option is C $\frac{3}{2}$

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

$={lim}_{h\to 0}\left(\frac{f^{\prime }(2h+2+h^2)(2+2h)}{f^{\prime }(h+1-h^2)(1-2h)}\right)$ using L'Hospitals rule

$=\left(\frac{f^{\prime }(0+2+0)(2+0)}{f^{\prime }(0+1-0)(1-0)}\right)$

$=\left(\frac{f^{\prime }\left(2\right)\left(2\right)}{f^{\prime }\left(1\right)\left(1\right)}\right)$ from above values $f^{\prime }\left(2\right)=6$ and $f^{\prime }\left(1\right)=4$ we have

$=\frac{6\times 2}{4\times 1}=\frac{3}{2}$