# Evaluate Limit: $\frac{0}{0}$ - Can Someone Help?

• MHB
• joypav
In summary, the limit as x approaches 0 from the positive side of (e^(-1/x))/(x^n) is equal to 0. The process involves using L'Hospital's rule and transforming the expression into the form of (1-n*log(u))/ (1/u), where u approaches infinity. This simplifies to e^(-∞) which is equal to 0.
joypav

I'm trying to show...

$\lim_{{x}\to{0^+}}\left(\frac{e^(\frac{-1}{x})}{x^n}\right)=0$

I guess my calculus is a bit rusty. Can someone help me out?
Here's what I've got.

$\lim_{{x}\to{0^+}}ln\left(\frac{e^(\frac{-1}{x})}{x^n}\right)=\lim_{{x}\to{0^+}}\left(-\frac{1}{x}-ln(x^n)\right)=-\infty+\infty$

So, I put it in a form to get $\frac{0}{0}$ so I can use L'Hospital's.

$\lim_{{x}\to{0^+}}\frac{x+\frac{1}{ln(x^n)}}{-\frac{x}{ln(x^n)}}$

Is this how I am supposed to be proceeding? (Wondering)

There's a trick to this one.

-(1/x + nlog(x))

-(1/x - nlog(1/x))

-(u - nlog(u)) as u approaches infinity (a swap with 1/x) :)

-(1 - nlog(u)/u)/(1/u)

Hence we have (if you'll pardon the notation ) $e^{-\infty}=0$.

I'm assuming $n$ is a constant.

$-(u - nlog(u))$

$\frac{-(1-nlog(u)}{\frac{1}{u}}$

How do you get from here to here?
I understand that if you factor out a $u$ the first term becomes $1$, but why does $nlog(u)$ stay the same?

My apologies - i made an error. I have edited my post.

greg1313 said:
My apologies - i made an error. I have edited my post.

Gotcha. Thanks!

## 1. What does it mean when a limit evaluates to $\frac{0}{0}$?

When a limit evaluates to $\frac{0}{0}$, it means that the function being evaluated has a point where both the numerator and denominator approach 0. This is known as an indeterminate form and requires further analysis to determine the actual limit value.

## 2. How do you solve a limit that evaluates to $\frac{0}{0}$?

To solve a limit that evaluates to $\frac{0}{0}$, you must use a mathematical technique called L'Hopital's rule. This rule allows you to take the derivative of both the numerator and denominator separately and then evaluate the limit again. If the new limit still evaluates to $\frac{0}{0}$, you can repeat the process until you get a non-indeterminate form.

## 3. Can you give an example of a limit that evaluates to $\frac{0}{0}$?

One example of a limit that evaluates to $\frac{0}{0}$ is $\lim_{x \to 0} \frac{x}{x}$. As x approaches 0, both the numerator and denominator become 0, resulting in an indeterminate form. Using L'Hopital's rule, we can take the derivative of the top and bottom, giving us $\lim_{x \to 0} \frac{1}{1}$, which evaluates to 1.

## 4. Why is it important to understand limits that evaluate to $\frac{0}{0}$?

Understanding limits that evaluate to $\frac{0}{0}$ is important because they often occur in real-world applications and are essential in determining the behavior of a function at a specific point. Additionally, these limits are crucial in calculus and are the basis for many other mathematical concepts.

## 5. Are there any other indeterminate forms besides $\frac{0}{0}$?

Yes, there are two other indeterminate forms: $\frac{\infty}{\infty}$ and $0 \cdot \infty$. These forms also require further analysis to determine the actual limit value, and L'Hopital's rule can be used to solve them as well.

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