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Real Inequality/Limit

How to prove that as x approaches infinity, xex\frac{x}{e^x} approaches 0. It seems so simple yet I am unable to come up with a formal proof. I would appreciate proofs without L'hospital rule or any complicated calculus.

Does the same way work for x2ex\frac{x^2}{e^x}?
Reply 1
Original post by esrever
How to prove that as x approaches infinity, xex\frac{x}{e^x} approaches 0. It seems so simple yet I am unable to come up with a formal proof. I would appreciate proofs without L'hospital rule or any complicated calculus.

Does the same way work for x2ex\frac{x^2}{e^x}?


You'd probably need to use the definition of
e^x = (1 + x/n)^n
as n tends to infinity, then use Bernoulli's inequality. Have a go.

Yes to the second part.
Reply 2
Original post by mqb2766
You'd probably need to use the definition of
e^x = (1 + x/n)^n
as n tends to infinity, then use Bernoulli's inequality. Have a go.

Yes to the second part.


Thanks :biggrin:. But how would you go about x/2^x. Should I just write 2 as eln2e^{\ln 2}?
Reply 3
Original post by esrever
Thanks :biggrin:. But how would you go about x/2^x. Should I just write 2 as eln2e^{\ln 2}?

Couple of ways you could do it but probably the easiest is something like x = 2z
x^2/e^x = 4z^2 / (e^z)^2 = 4*(z/e^z)^2
So if the first one works, so does the 2nd.
Original post by esrever
How to prove that as x approaches infinity, xex\frac{x}{e^x} approaches 0. It seems so simple yet I am unable to come up with a formal proof. I would appreciate proofs without L'hospital rule or any complicated calculus.

Does the same way work for x2ex\frac{x^2}{e^x}?

@mqb2766 has already given you a good suggestion. Here's another that you can use if you're allowed to assume the series expansion of e^x.

Look at each of the terms of the expansion of e^x; what can you say about them? Therefore what can you say about the relationship between any one term of the expansion and e^x itself? Now choose the appropriate term! This approach works for any value of n in xnexx^n e^{-x}.
Reply 5
Original post by esrever
How to prove that as x approaches infinity, xex\frac{x}{e^x} approaches 0. It seems so simple yet I am unable to come up with a formal proof. I would appreciate proofs without L'hospital rule or any complicated calculus.

Does the same way work for x2ex\frac{x^2}{e^x}?

As x goes to infinity ex=1+x+x22+ e^x = 1 + x + \dfrac{x^2}{2} + \dots , so 1ex<2x2 \dfrac{1}{e^x} < \dfrac{2}{x^2} then replace e^x in your expression by this to get an upper bound which should go to 0
For xR+x \in \mathbb{R}^{+}, we have 0xex11+x0\displaystyle 0 \le \frac{x}{e^x} \le \frac{1}{1+x} \to 0 and 0x2ex2x1+x20\displaystyle 0 \le \frac{x^2}{e^x} \le \frac{2x}{1+x^2} \to 0.

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