# limit = e

• December 8th 2008, 02:21 AM
Greengoblin
limit = e
Can someone show me how:

$\lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$
• December 8th 2008, 02:26 AM
Jameson
Quote:

Originally Posted by Greengoblin
Can someone show me how:

$\lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$

Say that this limit exists and it equals A.

Then $=\frac{1}{t}\ln(1+t)=\ln(A)$

The LHS is also $\frac{\ln(1+t)}{t}$ and when evaluating at t=0 you get 0/0, which is indeterminate and L'Hopitals can be used.

So $\ln(A)=\frac{\frac{1}{1+t}}{1}$

Try the limit again, and use algebra to solve for A again.

edit: This is in the precalc section, so forgive me for using calculus.
• December 8th 2008, 02:30 AM
Chop Suey
Quote:

Originally Posted by Greengoblin
Can someone show me how:

$\lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$

Another way is to sub $u=\frac{1}{t}$ so that $u \to \infty~\text{as}~t \to 0$ to get the famous limit:

$\lim_{u \to \infty} \left(1+\frac{1}{u}\right)^u = e$

@Jameson: I am sure that Greengoblin's level is beyond precal. ;)
• December 8th 2008, 02:35 AM
Jameson
Yes that would appear so after looking some more :)
• December 8th 2008, 03:57 AM
Greengoblin
Thank you both so much, I understand it now. :)
• December 8th 2008, 09:24 AM
Jhevon
Alternate solution:

recall that $e^{\ln X} = X$ for any $X > 0$.

so that:

$\lim_{t \to 0}(1 + t)^{\frac 1t} = \lim_{t \to 0}\text{exp} \left( \frac 1t \ln (1 + t) \right)$

$= \text{exp} \left( \lim_{t \to 0} \frac {\ln (1 + t)}t \right)$

now since $\lim_{t \to 0} \frac {\ln (1 + t)}t = 1$ (By L'Hopital's rule, or power series or ...)

we have that the limit is $\text{exp}(1) = e$

This is very similar to Jameson's method, both of these methods you should consider when taking the limit of an expression with the variable in the power.

That is, let $A$ and $B$ be expressions in a single variable in which the variable appears, then we can usually find various limits that give indeterminate forms by doing the following:

The method Jameson used:

Say $y = A^B$

Then $\ln y = B \ln A$

so that $\lim \ln y = \lim B \ln A$

and so $\lim y = e^{\lim B \ln A}$ by taking the anti-log of both sides.

The method I used:

simply write $y = A^B$ as $e^{\ln A^B} = e^{B \ln A}$, and take the limit of both sides.

note that $\lim e^X = e^{\lim X}$
• December 8th 2008, 09:43 AM
Greengoblin
Thanks, but could you explain how $\lim e^X=e^{\lim X}$ ?
• December 8th 2008, 09:45 AM
Chop Suey
If $\lim_{x \to a} f(x) = L$ and $g$ is continuous on $L$, then $\lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a} f(x)\right) = g(L)$
• December 8th 2008, 09:48 AM
Jhevon
Quote:

Originally Posted by Greengoblin
Thanks, but could you explain how $\lim e^X=e^{\lim X}$ ?

the rigorous theory might get a bit complicated, but it is somewhat similar to the fact that we can factor constants out side of limits. since the limit deals with X and not e, taking the limit of e to some power really only affects the power, since anything that is not a variable in X is not affected by the limit.

EDIT: Chop Suey was a bit more rigorous :p.
• December 9th 2008, 02:59 AM
Greengoblin
Thanks, I got it now. ;) I need to have a good look at limits again, as I haven't looked at them for ages and have forgotten alot.