Give the Limit

1. Mar 31, 2009

intenzxboi

1. The problem statement, all variables and given/known data

give the limit if it exist of (1-1/n)^n n=1 to infinity

the teacher wrote:
lim n goes to infinity (1+x/n)^n = e^x

so therefore
lim n goes to infinity (1-1/n)^n = e^-1

can someone explain this to me.

2. Mar 31, 2009

rwisz

All they've done is given x a value. And since the two expressions are equivalent, when x is a specific value in one expression, to remain equivalent x has the same value in the other.

So all your teacher did was substitute -1 for x?

3. Mar 31, 2009

intenzxboi

how i'm i suppose to know that

(1+x/n)^n = e^x ??

4. Mar 31, 2009

rwisz

Ah, yes I was wondering if that was your question.

Well the proof exists online... it's not too terribly difficult.

To prove that (where x is some constant) $$\mathop{\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n} = e^x$$

We first set the limit equal to an arbitrary variable (y for instance).

$$y =\mathop {\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n}$$

Next we'll utilize the rules of the natural logarithm to bring down the power from the quantity on the right.

$$lny =\mathop {\lim }\limits_{n \to \infty } n ln(1+\frac{x}{n})$$

Now to simplify and move forward, we must do some algebraic manipulation.

(Multiply by 1, which in this case we represent with $$\frac{\frac{1}{n}}{\frac{1}{n}}$$)

$$lny =\mathop {\lim }\limits_{n \to \infty }n ln(1+\frac{x}{n})*\frac{\frac{1}{n}}{\frac{1}{n}}$$

As you can see, this cancels out the n in the numerator, AND gives us a denominator.

$$lny =\mathop {\lim }\limits_{n \to \infty } \frac{ln(1+\frac{x}{n})}{\frac{1}{n}}$$

At this point we would re-evaluate the limit, and it would return 0/0 (plug in infinity, it will give 0/0).

So, we use L'Hopital's Limit Rule to take the derivative of the numerator and denominator, respectively.
Afterwords you should get,

$$x*\frac{\frac{n^{-2}}{1+\frac{1}{n}}}{n^{-2}}$$ Which you can see would cancel out the $$n^-2$$'s from the expression. Which would yield:

$$lny =\mathop {\lim }\limits_{n \to \infty }\frac{x}{1+\frac{1}{n}}$$

Evaluate the limit of the right side, and you'd get $$\frac{x}{1+0}$$ or simply $$x$$

So you end up with $$lny = x$$ after which you can exponentiate both sides, and this gives you your final answer!

$$y = e^x$$

And since y = both your original limit and your new expression e^x...

$$\mathop{\lim }\limits_{n \to \infty } (1+\frac{x}{n})^{n} = e^x$$

Now there's a few steps taken out of the explanation, such as actually taking the derivatives of L'Hopital's rule... but I would assume you could perform those by yourself.

I hope this helps!

Last edited: Mar 31, 2009
5. Apr 1, 2009

HallsofIvy

Staff Emeritus
What is your definition of "e"?