0
$\begingroup$

How could I show that the following limit tends to $0$? The constant $c$ can take any value strictly greater than $1.$

$$\lim_{n \to \infty}\dfrac{e^\sqrt{n}}{c^n}$$

I'm having trouble on how to approach this problem.

$\endgroup$
  • $\begingroup$ Sketch a graph of the logarithm of your ratio on the vertical axis and $\sqrt n$ on the horizontal axis. You will meet an old friend. $\endgroup$ – kimchi lover Feb 3 at 12:11
  • $\begingroup$ The limit is equal to zero $\endgroup$ – Dr. Sonnhard Graubner Feb 3 at 12:16
2
$\begingroup$

Your limit can be write as $\lim e^{\sqrt{n}-log(c)n}$, and $\lim (\sqrt{n}-log(c)n)= -\infty$, so $\lim e^{\sqrt{n}-log(c)n}=0$ (or you can make a rigorous proof).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.