Limit question (e^x)/x as x->inf without using calculus
micha...@aol.com
shorted out. He is doing limits and has a problem where he was asked
to find the limits of (e^x)/x as x->inf. He has not even done any
derivatives yet so I can not use l'hopitals rule. Can you guys suggest
an approach to finding and proving this limit. By l'hopital and taylor
expansion and general intuition I know it must go to infinity but I
don't know how to prove that without calculus. I have tried various
ways of taking the log and doing some algebra but then i have to
assume that X grows faster than log(X) and I am not sure how to prove
that without calculus.
Any ideas?
Richard E. Harke
I thought the exponential was defined by its power series,
no need to invoke derivatives. If that is not satisfactory,
go back to the definition limit n->inf (1+x/n)^n
Richard
achille
How about employ some form of ratio test (assume x >= 1):
(e^(x+1)/(x+1)) / (e^x/x) = e x /(1+x) >= e/2.
=> e^x/x >= (e/2)^(floor(x)-1)*(e^1/1) = 2 * (e/2)^floor(x)
Zdislav V. Kovarik
***********
Bernoulli inequality is as pre-calculus as it can get (classical proof
goes by induction):
(1+b)^m > 1+bm (m >=2 integer, b>-1, b nonzero)
**********
Now:
For a>1 (such as a=e), write a = (1+b)^2, where b=sqrt(a)-1 > 0.
Consider x > 3, and denote y = floor(x), so
y is integer, x-1 < y <= x
a^x = ((1+b)^x)^2 >= ((1+b)^y)^2 > (1+b*y)^2 > (1+b*(x-1))^2
a^x / x > (1+b*(x-1))^2/x = b^2*x + (something bounded).
Messy, but elementary.
Cheers,
ZVK(Slavek).
-------------------------------------
Based on a children's riddle:
Q: How do you catch an elephant?
A: Catch two and let one go.
David C. Ullrich
You can give various "proofs". But you can't _prove_ this
wirhout some calculus, because you can't even _define_
e^x in the first place without calculus!
>Any ideas?
Dave L. Renfro
> Hi I am tutoring a calculus student and apparently my
> brain has shorted out. He is doing limits and has a
> problem where he was asked to find the limits of
> (e^x)/x as x->inf. He has not even done any derivatives
> yet so I can not use l'hopitals rule. Can you guys
> suggest an approach to finding and proving this limit.
As David C. Ullrich pointed out, I think you're going
to encounter a lot of difficulty in trying to _prove_
the limit is infinity at the beginning of a calculus
course, at least by using a proof that would be
understandable to most of the students. An approach
I've used before is to look at what happens if we
let x = e^10, e^100, e^100, ...
More precisely, let y = e^x. Then x --> infinity
if and only if y --> infinity, so we can reformulate
the limit as
exp(e^y) / e^y as y --> infinity,
or as
exp(e^y - y) as y --> infinity.
Now it's just a matter of observing that e^y - y
approaches infinity as y --> infinity, which will
seem much more convincing at this level than the
original limit (although a true rigorous proof is
probably just as hard to attain). Note that e^y > 2^y,
so it suffices to show 2^y - y approaches infinity
as y --> infinity, and 2^y - y --> infinity is likely
to be much easier for the student to contemplate
than e^y - y --> infinity.
Dave L. Renfro
Anon
If h is a positive number and n is a positive integer, then
(1 + h)^n >= 1 + nh.
This is a consequence of the binomial theorem.
For an arbitrary number a greater than one, the quotient a^x/x tends
to infinity as x increases. Proof:
Let sqrt(a) = b = 1 + h, so b > 1 and h > 0. Let n be the integer
such that n <= x < n+1. Take x > 1 so that n >= 1.
sqrt(a^x/x) = b^x/sqrt(x) = (1+h)^x/sqrt(x) > (1+h)^n/sqrt(n+1) >
(1+nh)/sqrt(n+1) > nh/sqrt(2n) = h/sqrt(2) * sqrt(n)
so that a^x/x > h^2/2 * n, which therefore tends to infinity with x.
William Elliot
> Hi I am tutoring a calculus student and apparently my brain has
> shorted out. He is doing limits and has a problem where he was asked
> to find the limits of (e^x)/x as x->inf. He has not even done any
> derivatives yet so I can not use l'hopitals rule. Can you guys suggest
> an approach to finding and proving this limit.
Using for all x >= 1, f(x) = e^x / x > 1:
lim(x->oo) e^x / x = lim(x->oo) e^2x / 2x
= lim(x->oo) e^x/2 * e^x / x >= lim(x->oo) e^x / 2 = oo.
mjc
If f(x) = e^x/x, where x >= 1, then
f(2x) = e^(2x)/(2x) = f(x)e^x/2 >= f(x)(e/2).
By induction, f(2^k x) >= f(x)(e/2)^k.