3
$\begingroup$

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,

$\endgroup$
2
  • 1
    $\begingroup$ For the solution with L'Hospital's rule, you should first try to calculate it yourself! $\endgroup$
    – 5xum
    Mar 17, 2016 at 9:50
  • $\begingroup$ @5xum I tried with L'Hospital's rule but the denominator keeps growing ... Have you seen this problem solved using this technique? $\endgroup$
    – texmex
    Mar 17, 2016 at 10:13

3 Answers 3

Reset to default
4
$\begingroup$

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, $$ \lim_{x\to \infty} \frac{e^x}{e^{x^2}} = \lim_{y \to -\infty} e^y = 0 $$

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, $x-x^2 \to -\infty$.

$\endgroup$
5
  • $\begingroup$ could you please clarify how $x-x^{2}$ tends to negative infinity ? $\endgroup$
    – texmex
    Mar 17, 2016 at 10:09
  • $\begingroup$ @user249613 Added something. $\endgroup$
    – martini
    Mar 17, 2016 at 10:13
  • $\begingroup$ what a wonderful proof .... Much appreciated thanks $\endgroup$
    – texmex
    Mar 17, 2016 at 10:23
  • $\begingroup$ One simple clarification, do we take $x \ge \max\{1,L+1\}$ since $x$ tends to infinity? $\endgroup$
    – texmex
    Mar 17, 2016 at 10:25
  • $\begingroup$ Right. $\lim_{x\to\infty} (x-x^2) = -\infty$ means that $$ \forall L \ge 0\, \exists x_0 \, \forall x \ge x_0: x-x^2 \le -L $$ we have $x_0 = \max\{1, L+1\}$ here. $\endgroup$
    – martini
    Mar 17, 2016 at 10:26
2
$\begingroup$

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 $x \to \infty$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .