Can someone show me how:

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

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- Dec 8th 2008, 01:21 AMGreengoblinlimit = e
Can someone show me how:

$\displaystyle \lim_{t\to 0} (1+t)^{\frac{1}{t}}=e$ - Dec 8th 2008, 01:26 AMJameson
Say that this limit exists and it equals A.

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

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

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

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. - Dec 8th 2008, 01:30 AMChop Suey
Another way is to sub $\displaystyle u=\frac{1}{t}$ so that $\displaystyle u \to \infty~\text{as}~t \to 0$ to get the famous limit:

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

@Jameson: I am sure that Greengoblin's level is beyond precal. ;) - Dec 8th 2008, 01:35 AMJameson
Yes that would appear so after looking some more :)

- Dec 8th 2008, 02:57 AMGreengoblin
Thank you both so much, I understand it now. :)

- Dec 8th 2008, 08:24 AMJhevon
Alternate solution:

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

so that:

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

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

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

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

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 $\displaystyle A $ and $\displaystyle B$ 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 $\displaystyle y = A^B$

Then $\displaystyle \ln y = B \ln A$

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

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

The method I used:

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

note that $\displaystyle \lim e^X = e^{\lim X}$ - Dec 8th 2008, 08:43 AMGreengoblin
Thanks, but could you explain how $\displaystyle \lim e^X=e^{\lim X}$ ?

- Dec 8th 2008, 08:45 AMChop Suey
If $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle g$ is continuous on $\displaystyle L$, then $\displaystyle \lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a} f(x)\right) = g(L)$

- Dec 8th 2008, 08:48 AMJhevon
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. - Dec 9th 2008, 01:59 AMGreengoblin
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.