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.

  • $\begingroup$ 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

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).


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