2016/100) find $lim_{ x-> inf} (\frac{x+6}{x+1})^{x+4}$

we have $lim_{ x-> inf}(\frac{x+6}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+1} (1+ \frac{5}{x+1})^3$
= $e^5 * 1 = e^5$