Let f:R (0,) be a twice differentiable function such that f(3)=18,f^'(3)=0 and f^''(3)=4.

Question

Let $f:R	o (0,\infty)$ be a twice differentiable function such that $f(3)=18,{f}^{'}(3)=0$ and ${f}^{''}(3)=4$.

Then $\underset{x	o 1}{lim}({\log }_{e}{(\frac{f(2+x)}{f(3)})}^{\frac{18}{(x-1{)}^{2}}})$ is equal to :

Accepted Answer

The bracket is a $1^\infty$ form since $f(x+2)/f(3)\to f(3)/f(3)=1$ as $x\to1$. Taking the log converts the exponent limit to
$$\lim_{x\to1}\frac{18}{(x-1)^2}\left(\frac{f(x+2)}{f(3)}-1\right)$$
which is a $\frac00$ form (numerator $\to0$ since $f(1+2)=f(3)$). Applying L'Hopital's rule twice (differentiating with respect to $x$):

$$=\lim_{x\to1}\frac{18\,f'(x+2)}{2(x-1)f(3)}=\lim_{x\to1}\frac{18\,f''(x+2)}{2f(3)}=\frac{18f''(3)}{2f(3)}=\frac{18\cdot4}{2\cdot18}=2$$

Solutions · 1