Here's a math question: what is the limit X->0+ of e-(1+x)^(1/x)/x?

Answer 1

ok...well 1/x becomes basically infinity as x gets really close to zero...so (1/x)/x woud be infinity/something really close to 0 which would make the value infinity

so we have

e-(1-x)^infinity
well...thats ok

e-(1 - (basically zero))^infinity is
e-1^infinity
which is
e-1

so the limit as x approachs zero from the right is e-1 which is approximately 1.71828

hope this helps
matttlocke

Answer 2

I'm rewriting this and asking if this is what you meant:

e - [ (1+x)^(1/x)]/x limit as x goes to 0?

or

[e - (1+x)^(1/x) ] /x

I would like to mention that e = lim x->inf (1 + 1/x)^x

So, you could replace x with 1/y and take the lim as y->inf

You still get a limit of the form 0/0 and need to use L'H

Using L'Hospitol's on this and taking the D of top and bottom:

-1/x (1 + x)^(1/x - 1) (-1/x^2)
------------------------------------
1

This rewrites to:

(1 + x)^(1/x -1)
-------------------------
x^3

The limit of the top is e.
The limit of the bottom is 0.

The limit of the original function is infinity.

By the way, graphing this doesn't do much good unless you replace x with 1/x and zoom way, way out or use a table. Your instructor is playing games with you and thinking you might just graph it and look at it and see how flat it becomes in the 10^(-10) range. However, it starts taking off right after that and appears to become unbounded.

Answer 3

with the help of enhance, e^x = a million + x/a million! + x^2/2! + x^3/3! + x^4/4! + ..... + ? ==> e^x - a million = a million + x/a million! + x^2/2! + x^3/3! + x^4/4! + ..... + ? - a million ==> e^x - a million = x/a million! + x^2/2! + x^3/3! + x^4/4! + ..... + ? [First term a million and -a million cancels] ==> [e^x - a million]/x = a million + x/2! + x^2/3! + x^3/4! + .... + ? [On dividing each and each term with the help of 'x', first term is a million and all different words have 'x' in numerator] for this reason as x -> 0, all words = 0, yet for first term. subsequently, the shrink of x->.0 [e^x-a million]/x = a million

Answer 4

For now forget the e, you want to figure lim (1 + x)^(1/x)/x
lim (1 + x)^(1/x) : look at the log: 1/xlog(1 + x) as x goes to 0 == (1/(1 + x))/1 = 1/(1 + x) => 1

Now raise e to this power (to undo the log we just took) so you have

lim x=>0 (1 + x)^(1/x) = e.


Put you don't just have this term, you have it over x, so this is a constant divided by a variable going to 0, which has no limit (is unbounded), so e - it is unbounded too.

Answer 5

I believe the answer is e-1

if x >0 the the (1+0)is being raised to a power of 0

anything raised to the zero power is 1 therefore the lim as x>0 = e-1

oops I copied the terms wrong my colleage above is correct...

Answer 6

I think you mean

Lim x->0 (e-(1+x)^(1/x))/x

this is of the form 0/0

use l hospital's rule
numerator

d/dx(e-(1+x)^(1/x))
= - d/dx(1+x)^(1/x)

y = (1+x)^(1/x)
ln y = 1/x(ln(1+x)
dy/dx/y = 1/x. 1/(1+x) + ln(1+x)

when x = 0 limit dy/dx = y(1+in(1+x) = y

denominatroe 1
so limt y = e

so limit of origninal function (e-e)/1 = 0