# Thread: How am I looking at this Limit wrong?

1. ## How am I looking at this Limit wrong?

Here's the limit I'm working with:

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

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

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 $\displaystyle\lim_{x\to\infty}(1)^x$, which is one.

So, my question is - how is the way I'm conceptually looking at this incorrect?

2. Use the substitution $t = \frac{2}{x} \implies t \to 0, x \to \infty$

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

Standard limit: $\lim_{t \to 0}(1+t)^{1/t}=e$

3. 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.

4. 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

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

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

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

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

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

Now by L'hopitals rule we get

$\exp\left( \lim_{x \to \infty} \frac{\frac{1}{1+\frac{2}{x}}\cdot \frac{-2}{x^2}}{\frac{-1}{x^2}} \right)=\exp\left( \lim_{x \to \infty} \frac{2}{1+\frac{2}{x}}\right)=e^2$

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

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.

6. Originally Posted by TheEmptySet

You have an indeterminate form!

What makes the original limit indeterminate?

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

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.

8. 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

9. 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.
$1^{\infty}$ is an indeterminate form.

10. 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.