I am trying to solve a limit and I need to understand which grows faster between $x^{10}$ and $e^{x-2}$.
I know that $\lim_{x \to \infty}\frac{e^x}{x^a}=\infty$ but how can I prove that $\lim_{x \to \infty}\frac{e^{x-2}}{x^a}=\infty$.I would go for a substitution where $t=x-2$, but I am not sure about it. Plus, what happens if there is a constant? How can I prove that for some $k>0$, $\lim_{x \to \infty}\frac{e^{x-2}}{Kx^a}=0$, that is for some constant this quantity goes to 0. Maybe this cannot happen, but, even in this case, how could I prove so? Thank you for your time.
Best Answer
Note that $$\lim_{x \to \infty} \frac{e^{x-2}}{x^a} = \frac{1}{e^2}\lim_{x \to \infty} \frac{e^{x}}{x^a} = \infty$$ Again $$\lim_{x \to \infty} \frac{e^{x}}{Kx^a} = \frac{1}{K}\lim_{x \to \infty} \frac{e^{x}}{x^a} = \infty$$
Just rememeber that the exponential function grows faster than $x^a$ for any $a$. A more intuitive way to see this is to note that $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$ So, no matter what $a$ you take, just take any integer let's say $n>a$ and note that $\frac{x^n}{n!}$ is in the series expansion of $e^x$.