Limit problem involving indeterminable powers

**auslmar** · June 14th 2008, 05:54 PM

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.

Thanks for your consideration,

Austin Martin

**Mathstud28** · June 14th 2008, 06:02 PM

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.

Thanks for your consideration,

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

**hatsoff** · June 14th 2008, 06:06 PM

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?

**mr fantastic** · June 14th 2008, 06:22 PM

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 $\lim_{n \rightarrow + \infty} \left(1 + \frac{2}{5n} \right)^n = \lim_{n \rightarrow + \infty} \left(1 + \frac{2/5}{n} \right)^n = e^{2/5}$ .

**Moo** · June 15th 2008, 01:04 AM

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

...

**auslmar** · June 15th 2008, 01:44 PM

Thanks to everyone, it really helped me out.

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