Aplicações e definições

Indeterminate forms also include $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, and $\infty^0$. But here, we will be concerned only with $\frac{0}{0}$. Ready?

So let $f(x)$ and $g(x)$ be two functions such that $\displaystyle\lim_{x \to a} f(x) = \displaystyle\lim_{x \to a} g(x) = 0$. Now, let us say we are interested in finding $$\lim_{x \to a} \frac {f(x)}{g(x)}.$$ Substituting $x=a$ would give us the indeterminate form $\frac{0}{0}$. What do?

l'Hopital's Rule to the rescue! Clearly, we can just take the derivative of $f(x)$ and $g(x)$ and things should automatically fall into place again, right? Maybe not. It might be the case that even after we get from $\displaystyle\lim_{x \to a} \frac {f(x)}{g(x)}$ to $\displaystyle\lim_{x \to a} \frac {f'(x)}{g'(x)}$, we are still at an indeterminate form. And, of course, we can apply l'Hopital's rule again and keep taking derivatives and hope that, at some point, the process terminates and we get a determinate form.

Most of the time, this happens, but sometimes it doesn't. It is rare to find an example of a $\frac{0}{0}$ where l'Hopital's rule fails us, but there are definitely cases where indeterminate forms have to be dealt with some other way.

If you find such an example, feel free to edit this post!

Created by the Bernoulli brothers and named after Guillaume de l'Hospital, l'Hospital's Rule is a helpful tool when a limit of the form $\lim_{x\to a}\frac{f(x)}{g(x)}$ is indeterminate i.e., $\frac{\infty}{\infty}$, $\frac{0}{0}$, $\infty-\infty$, $1^\infty$, etc. A function that has indeterminate form does not necessarily lack a limit; it is just unable to be evaluated in this form.

l'Hospital's Rule tells you to take the derivative of a ratio of functions by deriving the numerator and deriving the denominator. The derivative functions will look different, but the limit will be the same.

$\lim_{x\to a}\frac{f(x)}{g(x)}$ = $\lim_{x\to a}\frac{f'(x)}{g'(x)}$

It is important to note taking the derivative in this way is different than the quotient rule which states $\frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{[(f'(x)g(x)-f(x)g'(x)]}{g(x)^2]}$

For example, take the function

$\lim_{x\to\infty}\frac{(x^2-12x-15)}{(4-11x)}$

If you plug in the limit value right now, the function will take the form of $\frac{\infty}{-\infty}$; This is an indeterminate value because the limit of $\infty$ divided by $-\infty$ cannot be evaluated.

Taking the derivative, our new function is

$\lim_{x\to\infty}\frac{2x-12}{-11}$

By replacing the ratio of functions with the ratio of the functions' derivatives, we were able to obtain more information about the limit by taking it out of indeterminate form and we are able to actually evaluate it.

(Video) L'Hospital's Rule for Indeterminate Products

by PatrickJMT

Here's a video by patrickJMT showing you how to use l'Hospital's rule to evaluate limits in the indeterminate form $0\cdot\pm\infty$.

The idea is we will have two functions being multiplied that, when taken the limits of, give us $0\cdot\infty$. We can rewrite these functions in the following ways: $$f\cdot g=\frac{f}{\frac1g}=\frac{g}{\frac1f}$$ Algebraically, it doesn't matter which you put in the numerator, but some problems might turn out easier with a certain fraction in the numerator.

The first example he solves is: $$\lim_{x\rightarrow0^+}\sqrt{x}\cdot\ln{x}$$ Notice if we plug in $x=0$ we get $0\cdot-\infty$. We have this indeterminate form so we need to rearrange into one of the two fractions above. If we put $\ln{x}$ in the numerator we get an easier problem so let's do that. $$\lim_{x\rightarrow0^+}\frac{\ln{x}}{\frac{1}{\sqrt x}}=\lim_{x\rightarrow0^+}\frac{\ln x}{x^{-\frac12}}$$ This is now in the form infinity over infinity. So let's use l'Hospital's rule and take the derivative of the numerator and denominator $$L'H=\lim_{x\rightarrow0^+}\frac{\frac1x}{-\frac12x^{-\frac32}}$$ Now we can clean this up with some good ol' algebra. $$\begin{align} \lim_{x\rightarrow0^+}\frac{\frac1x}{-\frac12x^{-\frac32}}&=-2\lim_{x\rightarrow0^+}\frac{x^{\frac32}}{x^1}\\ &=-2\lim_{x\rightarrow0^+}\sqrt x\\ &=-2\sqrt{\color{red}{(0)}}=0\\ \end{align}$$

So we can say: $$\boxed{\lim_{x\rightarrow0^+}\sqrt{x}\cdot\ln{x}=0}$$

The next problem he tackles is: $$\lim_{x\rightarrow0}\cot2x\cdot\sin6x$$ Notice if we plug in $x=0$ we get $\infty\cdot0$. So let's rearrange this by putting one over cotangent in the denominator. $$\lim_{x\rightarrow0}\frac{\sin6x}{\frac{1}{\cot2x}}=\lim_{x\rightarrow0}\frac{\sin6x}{\tan2x}$$ This is now in the form zero over zero. So let's use l'Hospital's rule and take the derivative of the numerator and denominator. $$\begin{align} L'H&=\lim_{x\rightarrow0}\frac{6\cos6x}{2\sec^2{2x}}\\ &=\frac{6\cdot1}{2\cdot1}=3 \end{align}$$ So we can say $$\boxed{\lim_{x\rightarrow0}\cot2x\cdot\sin6x=3}$$

(Video) L'Hospital's Rule for Indeterminate Quotients

by PatrickJMT

Here's a video by patrickJMT where he explains l'Hospital's rule and shows you how to use it to solve limits with indeterminate quotients.

Suppose you have a limit in the form $\lim\limits_{x\rightarrow a}\frac{f(x)}{g(x)}$ and when you evaluate the limit you get something like $\frac{\infty}{\infty}$ or $\frac00$. L'Hospital's Rule says that we can take the derivative of both the numerator and denominator separately. $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}{}=^{L'H}\lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}$$ Sometime's people put an L'H above the equal sign to indicate that l'Hospital's rule is being used (I can't because of the LATEX so it's above and to the side). It's also important to note that you're taking the derivative of $f(x)$ and $g(x)$ separately. You're not taking the derivative of the quotient of $\frac{f(x)}{g(x)}$.

Let's see an example to get a better idea of how it works. The first example is a limit you might remember seeing while back as an identity: $$\lim_{x\rightarrow0}\frac{\sin x}{x}$$ Before we didn't even bother to show the proof to the identity because it was long and complicated. But if you plug in $x=0$ you'll see we get $\frac00$ which means it's time to introduce this problem to our friend l'Hospital. The first step is to take the derivative of both the top and the bottom of the fraction. $$\lim_{x\rightarrow0}\frac{\sin x}{x}=^{L'H}\lim_{x\rightarrow0}\frac{\cos x}{1}=\frac11$$ Now if we plug $x=0$ into this we get $1$ as our answer. So: $$\boxed{\lim_{x\rightarrow0}\frac{\sin x}{x}=1}$$

The next example he looks at is: $$\lim_{x\rightarrow\infty}\frac{\ln x}{x}$$ When we plug in $x=\infty$ we get $\frac{\infty}{\infty}$ so we can use l'Hospital's rule here. $$\begin{align} L'H&=\lim_{x\rightarrow\infty}\frac{\frac1x}{1}\\ &=\lim_{x\rightarrow\infty}\frac1x=0 \end{align}$$ So: $$\boxed{\lim_{x\rightarrow\infty}\frac{\ln x}{x}=0}$$

What if we had a variation to this problem: $$\lim_{x\rightarrow0^+}\frac{\ln x}{x}$$ Well if we plug in $x=0$ we get $\frac{-\infty}{0}$. This isn't one of the forms in which we can use l'Hospital's rule. But notice that we're getting a very large negative number over a very small number. A large number divided by a small number is still a large number. So we can say: $$\boxed{\lim_{x\rightarrow0^+}\frac{\ln x}{x}=-\infty}$$

(Video) L'Hospital's Rule for Indeterminate Differences

by PatrickJMT

Here's a video by patrickJMT showing you how to use l'Hospital's rule to evaluate limits with indeterminate differences.

The first problem he looks at is: $$\lim_{x\rightarrow\infty}x^2-x$$ We get infinity minus infinity but the $x^2$ is approaching infinity faster than the $x$. So intuitively you might say that this limit is infinity. Well, you're right! But let's see why...using ALGEBRA! We can factor out an $x$. $$\lim_{x\rightarrow\infty}x^2-x=\lim_{x\rightarrow\infty}x(x-1)$$ Now if we plug in infinity we get $\infty\cdot\infty$ which equals infinity. So our intuition was correct. $$\boxed{\lim_{x\rightarrow\infty}x^2-x=\infty}$$

The next problem is: $$\lim_{x\rightarrow\infty}xe^{\frac1x}-x$$ Plugging in infinity gives us an indeterminate difference. So let's factor out an $x$ like last time. $$\lim_{x\rightarrow\infty}xe^{\frac1x}-x=\lim_{x\rightarrow\infty}x(e^{\frac1x}-1)$$ Now if we plug in infinity we get $\infty\cdot0$. Well, now we have an indeterminate product. Luckily, we know exactly how to solve this. $$\lim_{x\rightarrow\infty}x(e^{\frac1x}-1)=\lim_{x\rightarrow\infty}\frac{e^{\frac1x}-1}{\frac1x}$$ From here we can apply l'Hospital's rule. $$\begin{align} L'H&=\lim_{x\rightarrow\infty}\frac{e^{\frac1x}\cdot-x^{-2}}{-x^{-2}}\\ &=\lim_{x\rightarrow\infty}e^{\frac1x}\\ &=e^0=1 \end{align}$$ So finally, $$\boxed{\lim_{x\rightarrow\infty}xe^{\frac1x}-x=1}$$

The final example he solves is: $$\lim_{x\rightarrow1}\frac{1}{\ln x}-\frac{1}{x-1}$$ Plugging in infinity gives us $\infty-\infty$. So what do we do here? We can't factor something out. Well, when ever you have fractions it's always a good step to find like denominators and combine them. Let's try that: $$\begin{align} &=\lim_{x\rightarrow1}\frac{1}{\ln x}\color{red}{\cdot\frac{x-1}{x-1}}-\frac{1}{x-1}\color{red}{\cdot\frac{\ln x}{\ln x}}\\ &=\lim_{x\rightarrow1}\frac{x-1-\ln x}{\ln x(x-1)} \end{align}$$ Well would you look at that we now have an indeterminate quotient! And we all know who shows up to the party with those. $$L'H=\lim_{x\rightarrow1}\frac{1-\frac1x}{\frac1x(x-1)+\ln x}$$ Hmm, this is pretty gross looking. Let's clean it up with some $x$ on the top and some $x$ on the bottom. $$\begin{align} &\lim_{x\rightarrow1}\frac{1-\frac1x}{\frac1x(x-1)+\ln x}\color{red}{\cdot\frac xx}\\ &=\lim_{x\rightarrow1}\frac{x-1}{x-1+x\ln x} \end{align}$$ Sadly, if we plug in $x=1$ we get $\frac00$ so we're not out of the woods yet. It's time for a reunion episode with our man l'Hospital. $$\begin{align} L'H&=\lim_{x\rightarrow1}\frac{1}{1+\ln x+x\frac1x}\\ &=\lim_{x\rightarrow1}\frac{1}{1+\ln x+1}\\ &=\lim_{x\rightarrow1}\frac{1}{2+\ln x}\\ &=\frac{1}{2+\ln \color{red}{(0)}}=\frac12 \end{align}$$ So finally we get, $$\boxed{\lim_{x\rightarrow1}\frac{1}{\ln x}-\frac{1}{x-1}=\frac12}$$