limit question as x->0?

Question

lim (1+x)^(1/x)
x->0

thanks in advance

i thought 1....but if u graph it its more like 2.7..?

Answer 1

Instead of trying to find this limit directly, try to find the logarithm of the limit:

ln [x→0]lim (1+x)^(1/x)

Since the natural logarithm is continuous, we may move it inside the limit:

[x→0]lim ln (1+x)^(1/x)

Using the laws of logarithms:

[x→0]lim (1/x) ln (1+x)
[x→0]lim ln (1+x)/x

This is now of the form 0/0, so we may use L'hopital's rule:

[x→0]lim (1+x)/1

And this is fairly obviously:

1

But we aren't looking for the logarithm of the limit ,but the limit itself. However, now we know that:

ln [x→0]lim (1+x)^(1/x) = 1

So exponentiating both sides yields:

[x→0]lim (1+x)^(1/x) = e

And we are done.

Answer 2

That limit approaches e (Euler's number, the base of the natural log). It's one of the two equivalent limits I can think of that are commonly used for that purpose.

The limit I see more often to define e is the following:

lim (1 + (1/x))^x
x->infinity

That's more or less a "definition" of e. Replacing x with its reciprocal 1/x in your limit produces the second version, we just have to be careful about how we do it.

Since x is never actually equal to zero, we know that 1/x exists. Since we're taking a limit, there's nothing wrong with substituting 1/x for x, we just have to change the bound of the limit. Now we're letting 1/x to go 0, so x goes to infinity. That's how we get the second limit from the first.

Thus your limit converges to e.

For more information on e (Euler's number), see the link below.

Answer 3

in case you plug interior the fee 0 to the equation... It won't deliver approximately the indeterminate type 0/0... = (?(0) -a million) / (0-a million) = -a million/-a million lim x->0 is a million... yet once you decide on for to do the cancellation subject.. you are able to... = (?(x) -a million) / (x-a million) (improve the denominator = (?(x) - a million) / (?(x) - a million) (?(x) + a million) (Cancel out the ?(x) - a million... = a million / (?(x) + a million (replace...) = a million / (?(0) + a million) = a million / 0 + a million = a million / a million lim x->0 is a million comparable answer... that stands out as the main suitable answer... thank you for analyzing this.. God bless you... :)

Answer 4

If we let n = 1/x and x = 1/n, x>0, then as x->0 we have n->infinity and the original expression can be written as

lim (1 + 1/n)^n = e where e = 2.718281828459....
x->0

The original expression is often taken as the definition of the number e.

In Finance, these expressions usually pop up in an explanation of how to compute interest that is compounded continuously.

Answer 5

I don't remember how to show it, but is similar to

lim (1+1/n)^n
n->oo

And this is e ~= 2.71828

Maybe:

ln (1+x)^(1/x) = ln(1+x)/x

l'hospital:

(1/1+x)/1 = 1 (x-->0)

Then the limit is e^1 = e ???

Answer 6

put x=1/n
now x-->0 so n-->inf

lim (1+x)^(1/x)
x->0

=lim(1+1/n)^n
n-->inf

=e [ by theorem,we know]