# Thread: Limit problem involving indeterminable powers

1. ## Limit problem involving indeterminable powers

The problem is as follows:

$\lim_{x\to\infty}(1+\frac{2}{x})^\frac{x}{5}$

What I've done:

$\lim_{x\to\infty}(1+\frac{2}{x})^\frac{x}{5} =$ $e^{\lim_{x\to\infty}\frac{x}{5}ln(1+\frac{2}{x})}$

So

$\lim_{x\to\infty}\frac{x}{5}ln(1+\frac{2}{x}) = \lim_{x\to\infty}\frac{ln(1+\frac{2}{x})}{\frac{5} {x}}$

Using L'Hospital's Rule

$= \lim_{x\to\infty}\frac{\frac{-2}{x^2+2x}}{\frac{-5}{x^2}}$

This the point where I get confused. It seems like this limit is indeterminable and it seems like if you keep doing L'Hospital's Rule that it only gets more complex only requires more and more applications of L'Hospital's. What am I doing wrong? Do I need to keep going? If anyone can point out my errs and shed some light on the correct path, I'd greatly appreciation.

Austin Martin

2. Originally Posted by auslmar
The problem is as follows:

$\lim_{x\to\infty}(1+\frac{2}{x})^\frac{x}{5}$

What I've done:

$\lim_{x\to\infty}(1+\frac{2}{x})^\frac{x}{5} =$ $e^{\lim_{x\to\infty}\frac{x}{5}ln(1+\frac{2}{x})}$

So

$\lim_{x\to\infty}\frac{x}{5}ln(1+\frac{2}{x}) = \lim_{x\to\infty}\frac{ln(1+\frac{2}{x})}{\frac{5} {x}}$

Using L'Hospital's Rule

$= \lim_{x\to\infty}\frac{\frac{-2}{x^2+2x}}{\frac{-5}{x^2}}$

This the point where I get confused. It seems like this limit is indeterminable and it seems like if you keep doing L'Hospital's Rule that it only gets more complex only requires more and more applications of L'Hospital's. What am I doing wrong? Do I need to keep going? If anyone can point out my errs and shed some light on the correct path, I'd greatly appreciation.

Austin Martin
Let

$L=\lim_{x\to\infty}\bigg(1+\frac{a}{x}\bigg)^{\fra c{x}{b}}\Rightarrow{L^b=\lim_{x\to\infty}\bigg(1+\ frac{a}{x}\bigg)^x}$

So $\ln(L^b)=\lim_{x\to\infty}\ln\bigg(1+\frac{a}{x}\b igg)x$

Now let $\varphi=\frac{1}{x}$

So as $x\to\infty\Rightarrow\varphi\to{0}$

Giving us

$\lim_{\varphi\to{0}}\frac{\ln(1+a\varphi)}{\varphi }=\lim_{\varphi\to{0}}\frac{a\varphi-\frac{a^2\varphi^2}{2}-\cdots}{\varphi}=a$

This implies that

$\ln(L^b)=a\Rightarrow{L=e^{\frac{a}{b}}}$

$\therefore\lim_{x\to\infty}\bigg(1+\frac{a}{x}\big g)^{\frac{x}{b}}=e^{\frac{a}{b}}$

Now apply it

3. Originally Posted by auslmar
$= \lim_{x\to\infty}\frac{\frac{-2}{x^2+2x}}{\frac{-5}{x^2}}$
$\lim_{x\to\infty}[(\frac{-2}{x^2+2x})(\frac{x^2}{-5})]$

$\frac{2}{5}\lim_{x\to\infty}\frac{x^2}{x^2+2x}$

Does that help?

4. Originally Posted by auslmar
The problem is as follows:

$\lim_{x\to\infty}(1+\frac{2}{x})^\frac{x}{5}$

[snip]
Standard limit: $\lim_{n \rightarrow + \infty} \left(1 + \frac{a}{n} \right)^n = e^{a}$.

Let $\frac{x}{5} = n \Rightarrow x = 5n$ and your limit becomes $\lim_{n \rightarrow + \infty} \left(1 + \frac{2}{5n} \right)^n = \lim_{n \rightarrow + \infty} \left(1 + \frac{2/5}{n} \right)^n = e^{2/5}$.

5. Originally Posted by auslmar
The problem is as follows:

...
So

$\lim_{x\to\infty}\frac{x}{5}ln(1+\frac{2}{x}) = \lim_{x\to\infty}\frac{ln(1+\frac{2}{x})}{\frac{5} {x}}$

...
You can see that $\lim_{x \to \infty} \frac 2x=0$

So $\ln\left(1+\frac 2x\right) \sim \frac 2x \quad \text{when} \quad x \to \infty$

...

6. Thanks to everyone, it really helped me out.