Find $\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$. Apply l’Hopital’s rule as many times as necessary, verifying your results after each application
$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3}$
First both the numerator and denominator go to infinity in the limit, so we can apply L'Hopital's rule.
Thus:
$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2}$
Again, both numerator and denominator go to infinity so:
$\displaystyle \lim_{x\to\infty}\frac{2e^{2x}}{3(x+5)^2} = \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)}$
Again, both numerator and denominator go to infinity so:
$\displaystyle \lim_{x\to\infty}\frac{4e^{2x}}{6(x+5)} = \lim_{x\to\infty}\frac{8e^{2x}}{6}$
Finally, only the numerator goes to infinity and the denominator is a constant, so the net result is:
$\displaystyle \lim_{x\to\infty}\frac{e^{2x}}{(x+5)^3} = \lim_{x\to\infty}\frac{8e^{2x}}{6} \to + \infty$
or more precisely, the limit is undefined.
-Dan