Question

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

• A. does not exist
• B. is equal to $-3/2$
• C. is equal to $3/2$
• D. is equal to $3$

Answer 1

Answer: (D)

is equal to $3$

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

Apply L-Hospital Rule

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