# How to find the limit of (e^(4/x) - 2x)^(X/2) as x-> 0+

How to find the limit of (e^(4/x) - 2x)^(X/2) as x--> 0+

## Homework Statement

$$\mathop{\lim}\limits_{x \to 0+} (e^{4/x} -2x)^{x/2}$$

## Homework Equations

if lim ln f(x) = L then $$\mathop{\lim}\limits_{x \to 0+} e^{ln f(x)} = e^L$$

## The Attempt at a Solution

Not too sure what my first step is. If I just plug in, I get 1. I tried taking the ln of the function, but that gives me a non-indeterminant result 0/2. Is the idea to take the ln, then derive, AND THEN solve?

Mark44
Mentor

You can't just "plug in" 0 for x since e4/x is undefined. I would start by letting y = (e4/x - 2x)x/2, and then taking the ln of both sides.

Then take the limit as x --> 0+. Your book probably has an example of this technique.

VietDao29
Homework Helper

## Homework Statement

$$\mathop{\lim}\limits_{x \to 0+} (e^{4/x} -2x)^{x/2}$$

## Homework Equations

if lim ln f(x) = L then $$\mathop{\lim}\limits_{x \to 0+} e^{ln f(x)} = e^L$$

## The Attempt at a Solution

Not too sure what my first step is. If I just plug in, I get 1. I tried taking the ln of the function, but that gives me a non-indeterminant result 0/2. Is the idea to take the ln, then derive, AND THEN solve?

To make it a little bit clearer for you. I'll give you an example:
Example:
Evaluate:
$$\lim_{x \rightarrow 0 ^ {+}} x ^ x$$
This is the indeterminate form 00.

Note that: When encountering $$0 ^ 0$$, or $$\infty ^ 0$$, we often consider letting y = the expression we need to take the limit. When take ln of both sides. Then try to find the limit of ln(y). (this is what Mark44 told you).

Ok, so in my example. Let:
y = xx
ln(y) = xln(x)

So: $$\lim_{x \rightarrow 0 ^ {+}} \ln y = \lim_{x \rightarrow 0 ^ {+}} x \ln x$$

The RHS is in the indeterminate form $$0 \times \infty$$. However, we can only apply L'Hopital's Rule when it's $$\frac{0}{0}, \mbox{ or } \frac{\infty}{\infty}$$. So we do a little trick to change the indeterminate form, like this:

$$... = \lim_{x \rightarrow 0 ^ {+}} \frac{\ln x}{\frac{1}{x}}$$

Now, it's $$\frac{\infty}{\infty}$$. Applying L'Hopital Rule here, we have:
$$... = \lim_{x \rightarrow 0 ^ {+}} \frac{\frac{1}{x}}{-\frac{1}{x ^ 2}} = \lim_{x \rightarrow 0 ^ {+}} (-x) = 0$$.

By the definition of y, we have: $$\lim_{x \rightarrow 0 ^ {+}} \ln y = 0$$, which in turn means that: $$\lim_{x \rightarrow 0 ^ {+}} y = e ^ 0 = 1$$.

So: $$\lim_{x \rightarrow 0 ^ {+}} x ^ x = 1$$.

Now, let's see if you can tackle your problem. It's pretty much the same thing. :)