How am I looking at this Limit wrong?

• May 19th 2011, 12:27 PM
Lancet
How am I looking at this Limit wrong?
Here's the limit I'm working with:

$\displaystyle \displaystyle\lim_{x\to\infty}(1 + \frac{2}{x})^x$

Now, I understand that the correct answer is $\displaystyle e^2$. While I don't fully understand how you get there, I understand that you can manipulate this until you come up with $\displaystyle e^2$.

However, here's how I'm looking at it...

As x approaches infinity, the expression within the parentheses approaches 1. Which to me, means that the limit of this whole thing is $\displaystyle \displaystyle\lim_{x\to\infty}(1)^x$, which is one.

So, my question is - how is the way I'm conceptually looking at this incorrect?
• May 19th 2011, 12:39 PM
Mondreus
Use the substitution $\displaystyle t = \frac{2}{x} \implies t \to 0, x \to \infty$

$\displaystyle \lim_{t \to 0}(1+t)^{2/t}=\lim_{t \to 0}\left((1+t)^{1/t}\right)^2$

Standard limit: $\displaystyle \lim_{t \to 0}(1+t)^{1/t}=e$
• May 19th 2011, 12:53 PM
Lancet
Thanks, but that's just more manipulation (and it leaves me in the same boat, because I'm not sure what makes that last standard limit true). I was looking to understand what was wrong - conceptually - with the way I was looking at it.
• May 19th 2011, 01:04 PM
TheEmptySet
Quote:

Originally Posted by Lancet
Thanks, but that's just more manipulation (and it leaves me in the same boat, because I'm not sure what makes that last standard limit true). I was looking to understand what was wrong - conceptually - with the way I was looking at it.

You have an indeterminate form!

Just like

$\displaystyle \frac{0}{0} \text{ or } \frac{\infty}{\infty}$

So we need to resolve this limit in a different way

Since all of the functions are continuous and the natural log and exponential are monotone increasing we can use this identity

$\displaystyle \lim_{x \to \infty}\left( 1+\frac{2}{x}\right)^x=\exp\left( \lim_{x \to \infty}x \ln\left(1+\frac{2}{x} \right) \right)$

Now this is of the form zero times and infinity and can be put in the form 0 over 0

$\displaystyle \exp\left( \lim_{x \to \infty}\frac{\ln\left(1+\frac{2}{x} \right)}{\frac{1}{x}} \right)$

Now by L'hopitals rule we get

$\displaystyle \exp\left( \lim_{x \to \infty} \frac{\frac{1}{1+\frac{2}{x}}\cdot \frac{-2}{x^2}}{\frac{-1}{x^2}} \right)=\exp\left( \lim_{x \to \infty} \frac{2}{1+\frac{2}{x}}\right)=e^2$
• May 19th 2011, 01:14 PM
Plato
Quote:

Originally Posted by Lancet
Thanks, but that's just more manipulation (and it leaves me in the same boat, because I'm not sure what makes that last standard limit true). I was looking to understand what was wrong - conceptually - with the way I was looking at it.

Suppose that $\displaystyle a,~b,~\&~c$ are constants.
Then $\displaystyle \lim _{x \to \infty } \left( {1 + \frac{a}{{x + b}}} \right)^{cx} = e^{ac}$.

That is a very useful formula to know. It works is a wide range of this type question. The proof is not simply but the basic form is easy to remember.
• May 19th 2011, 01:14 PM
Lancet
Quote:

Originally Posted by TheEmptySet

You have an indeterminate form!

What makes the original limit indeterminate?
• May 19th 2011, 01:16 PM
Lancet
Quote:

Originally Posted by Plato
Suppose that $\displaystyle a,~b,~\&~c$ are constants.
Then $\displaystyle \lim _{x \to \infty } \left( {1 + \frac{a}{{x + b}}} \right)^{cx} = e^{ac}$.

That is a very useful formula to know. It works is a wide range of this type question. The proof is not simply but the basic form is easy to remember.

I appreciate you're trying, but all that is is a formula. It still doesn't explain what is wrong *conceptually* with how I was looking at the problem.
• May 19th 2011, 01:18 PM
TheEmptySet
Quote:

Originally Posted by Lancet
What makes the original limit indeterminate?

The base is approaching 1 but is always a little bigger than 1 and the exponent is going to infinity. Loosely speaking it is like a race does the base go to 1 faster then the exponent goes to infinity.

Indeterminate form - Wikipedia, the free encyclopedia
• May 19th 2011, 01:30 PM
Plato
Quote:

Originally Posted by Lancet
I appreciate you're trying, but all that is is a formula. It still doesn't explain what is wrong *conceptually* with how I was looking at the problem.

What wrong with your OP post is that it is an indeterminate form.
$\displaystyle 1^{\infty}$ is an indeterminate form.
• May 19th 2011, 02:27 PM
HallsofIvy
Quote:

Originally Posted by Lancet
I appreciate you're trying, but all that is is a formula. It still doesn't explain what is wrong *conceptually* with how I was looking at the problem.

Well, where exactly do you want to start? Do you accept that the derivative of e^x is e^x? Then you accept that lin_(x->0) (e^x- 1)/x= 1. That is, for x close to 0, (e^x- 1)/x is approximately 1 so e^x- 1 is approximately x and then e^x is approximately x+ 1. From that, x is approximately ln(x+ 1). From that ln(x+1)/x= ln(x+ 1)^{1/x} is approximately 1. Then (x+ 1)^{1/x} is approximately e.

Now, let n= 1/x so that x close to 0 becomes n very large: for very large n, (1+ (1/n))^n is approximately e and as n goes to infinity, lim_{n->infinity} (1+ (1/n))^n= e.