Help with upper-bounding $\lim_{x \rightarrow \infty} (\frac{c + x}{x}) ^ x$ with a constant

Question

I guess this is a pretty basic bound but I can't find it anywhere, I'm trying to upper-bound the term $(\frac{c + x}{x})^x$ when $c$ is a constant and $x$ is big. I get that $(\frac{c+x}{x})^x \approx 1^x$, but it's not clear whether it's upper-bounded by a constant. Can anyone help me?

Answer 1

Rewrite the limit as $\lim_{x\to \infty}(1+ \frac cx)^x$.

This should look familiar to you, as it's very close the definition of $e$. In fact, the answer is $e^c$. We can derive this fact as follows:

Let $y = \lim_{x\to \infty}(1+ \frac cx)^x$.

Then $\ln y = \lim_{x\to \infty} x \ln (1 + \frac cx)$ *

$= \lim_{x \to \infty}\frac{\ln (1 + \frac cx)}{\frac 1x}$

$= \lim_{x \to \infty} \frac{\frac{-c}{x^2 (1+ \frac cx)}}{\frac{-1}{x^2}}$**

$= \lim_{x \to \infty} \frac{c}{1+ \frac cx}$ ***

$= c$.

Then since that's only $\ln y$, and we want $y$, we set both sides as exponents of $e$.

So, $e^{\ln y} = e^c$. But $e^{\ln y} = y$, so $y = e^c$, as desired.

(*) follows by logarithm laws

(**) follows by L'Hopital's Rule

(***) follows by canceling out the $\frac{-1}{x^2}$ terms.

Hope this helped!