Limit of ratio of exponential function as variable tends to infinity:$\lim_{x\to \infty}\frac{e^x}{e^{x^2}}$

Question

Could someone please point out resources or provide the solution with detailed steps to calculate the following limits?

If possible, please provide solutions with or without using L'Hopital's rule.

$$ \underset{x\rightarrow\infty}{\lim}\left[\frac{e^{x}}{e^{x^{2}}}\right] $$

Thanks,

Answer 1

There is no need using l'Hospital here, for $$ \frac{e^x}{e^{x^2}} = e^{x-x^2} $$ And now $\lim_{x \to \infty} x-x^2 = -\infty$, that is, $$ \lim_{x\to \infty} \frac{e^x}{e^{x^2}} = \lim_{y \to -\infty} e^y = 0 $$

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Addendum: We will show that $\lim_{x\to \infty} (x-x^2) = -\infty$. For $x \ge 1$, we have $$ x-x^2 = x(1-x) \le 1-x $$ If now $L \ge 0$ is given, for $x \ge \max\{1,L+1\}$, we have $$ x-x^2 \le 1-x \le 1-(L+1) = -L $$ as $L$ was arbitrary, $x-x^2 \to -\infty$.

Answer 2

Hint: $$\frac{e^{x}}{e^{x^2}} = e^{x-x^2}$$

Answer 3

Maybe by writting $$\frac{e^{x}}{e^{x^2}} = e^{x-x^2}=e^{-x(x-1)}$$ you get that the limit is $0$ since $-x(x-1)\to -\infty$ as $x \to \infty$.