0
$\begingroup$

$\lim_{x\to\text{-}\infty} xe^x$. This is limit can be easily be seen that it approaches to 0 using graphs. But how to solve it algebraically?

$\endgroup$
  • 1
    $\begingroup$ Are you allowed to use that $te^{-t}\to 0$ as $t\to+\infty$? Or, what do you know about the exponential function? $\endgroup$ – mickep Sep 18 '15 at 12:12
  • 1
    $\begingroup$ Given that it is a transcendental function, and you are doing a limit, what do you mean "alegebraically"? And how is it linear? And what do you mean "behavior?" There are too many words in the title and many of them seem to have been arbitrarily chosen. $\endgroup$ – Thomas Andrews Sep 18 '15 at 12:13
  • $\begingroup$ i used a graph to see the limit. $\endgroup$ – Prachurjya Biswas Sep 18 '15 at 12:30
2
$\begingroup$

Elementary answer:
(I assume constantly that $x<0$.)
We can reformulate $$ xe^x = x\cdot 2^x \cdot \big(\frac e2\big)^x. $$ We know that $\lim_{x\to\ -\infty}\big(\frac e2\big)^x = 0$, as $\frac e2 > 1$ (we can use it, I suppose?), so it remains to show that $2^x x$ is bounded, which can be done, in fact, by induction.
Obviously, $2^x x < 0$. Also: $$ 2^x x \geq 2^{\lceil x\rceil}\! \lceil x\rceil = -\frac n {2^n} > -1. $$ where I set $n= -\lceil x \rceil$ and the inequality $n < 2^n$ can be proved by induction for every integer $n\geq 0$.
Of course, $2$ isn't in any way significant, anything in $(1,e)$ would do, but the induction proof of the inequality $n<2^n$ is quite standard, I think.

Note: you can also use the inequality $n^2 < e^n$ for every integer $n\geq 0$, also provable by induction, whence $x^2 e^x < 1$ for $x<0$ and, similarly to above: $$ 0 > x e^x = \frac 1x \cdot x^2 e^x > \frac 1x \stackrel{x\to -\infty}\longrightarrow 0. $$

$\endgroup$
  • $\begingroup$ thank you. but L'hospital rule would suffice $\endgroup$ – Prachurjya Biswas Sep 23 '15 at 12:28
1
$\begingroup$

You need to use L'Hospital's Rule. Start by writing $xe^x$ as $\frac{e^x}{1/x}$. See https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

$\endgroup$
  • $\begingroup$ It's correct, of course, but no need to use such a heavy machinery. See my answer. $\endgroup$ – eudes Sep 18 '15 at 14:22
  • $\begingroup$ tHank you . But the way L'hosputal rule is applied is different. $\endgroup$ – Prachurjya Biswas Sep 23 '15 at 12:28
  • $\begingroup$ Opps, sorry. It should have been $\frac{x}{e^{-x}}$ $\endgroup$ – 1-___- Sep 23 '15 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.