limit = e

**Greengoblin** · December 8th 2008, 01:21 AM

Can someone show me how:

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

**Jameson** · December 8th 2008, 01:26 AM

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.

**Chop Suey** · December 8th 2008, 01:30 AM

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.

**Jameson** · December 8th 2008, 01:35 AM

Yes that would appear so after looking some more

**Greengoblin** · December 8th 2008, 02:57 AM

Thank you both so much, I understand it now.

**Jhevon** · December 8th 2008, 08:24 AM

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 $\lim e^X = e^{\lim X}$

**Greengoblin** · December 8th 2008, 08:43 AM

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

**Chop Suey** · December 8th 2008, 08:45 AM

If $\lim_{x \to a} f(x) = L$ and $g$ is continuous on $L$ , then $\lim_{x \to a} g(f(x)) = g\left(\lim_{x \to a} f(x)\right) = g(L)$

**Jhevon** · December 8th 2008, 08:48 AM

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

.

**Greengoblin** · December 9th 2008, 01:59 AM

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.

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