Mathematics - Limits Question with Solution | TestHub

This limit is $1^{\infty }$ form

$\therefore \underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}{\left(\frac{1+f\left(3+x\right)-f\left(3\right)}{1+f\left(2-x\right)-f\left(2\right)}\right)}^{\frac{1}{x}}=e^{\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{x}\left(\frac{1+f\left(3+x\right)-f\left(3\right)}{1+f\left(2-x\right)-f\left(2\right)}-1\right)}$

$=e^{\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{x}\left(\frac{f\left(3+x\right)-f\left(3\right)-f\left(2-x\right)+f\left(2\right)}{1+f\left(2-x\right)-f\left(2\right)}\right)}\left(\frac{0}{0}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\right)$

$=e^{\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}} \frac{f^{\text{'}}\left(3+x\right)+f^{\text{'}}\left(2-x\right)}{x\left(-f^{\text{'}}\left(2-x\right)\right)+\left(1+f\left(2-x\right)-f\left(2\right)\right)}}$

$=e^0=1$.