# Indeterminate Forms and de L'hospital's Rule

## Key Questions

• L'hopital's rule is used primarily for finding the limit as $x \to a$ of a function of the form $f \frac{x}{g} \left(x\right)$, when the limits of f and g at a are such that $f \frac{a}{g} \left(a\right)$ results in an indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. In such cases, one can take the limit of the derivatives of those functions as $x \to a$. Thus, one would calculate ${\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$, which will be equal to the limit of the initial function.

As an example of a function where this may be useful, consider the function $\sin \frac{x}{x}$. In this case, $f \left(x\right) = \sin \left(x\right) , g \left(x\right) = x$. As $x \to 0$, $\sin \left(x\right) \to 0 \mathmr{and} x \to 0$. Thus,

lim_(x->0) sin(x)/x = 0/0 = ?

$\frac{0}{0}$ is an indeterminate form because we cannot precisely define what it is equal to.

However, by taking the derivatives, we find $f ' \left(x\right) = \cos \left(x\right) , g ' \left(x\right) = 1$. And thus...

${\lim}_{x \to 0} \sin \frac{x}{x} = {\lim}_{x \to 0} \cos \frac{x}{1} = {\lim}_{x \to 0} \cos \left(x\right) = \cos \left(0\right) = 1$

• l'Hopital's Rule occationally fails by falling into a never ending cycle. Let us look at the following limit.

${\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

by l'Hopital's Rule ($\infty$/$\infty$),

$= {\lim}_{n \to \infty} \frac{\frac{x}{\sqrt{{x}^{2} + 1}}}{1} = {\lim}_{n \to \infty} \frac{x}{\sqrt{{x}^{2} + 1}}$

by l'Hopital's Rule ($\infty$/$\infty$),

$= {\lim}_{n \to \infty} \frac{1}{\frac{x}{\sqrt{{x}^{2} + 1}}} = {\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

As you can see, the limit came back to the original limit after applying l'Hopital's Rule twice, which means that it will never yield a conclusion. So, we just need to try another approach.

${\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

by including the denominator under the square-root,

$= {\lim}_{n \to \infty} \sqrt{\frac{{x}^{2} + 1}{x} ^ 2}$

by simplifying the expression inside the square-root,

${\lim}_{n \to \infty} \sqrt{1 + \frac{1}{x} ^ 2} = \sqrt{1 + \frac{1}{\infty} ^ 2} = \sqrt{1 + 0} = 1$

So, we could come up with the limit without using l'Hopital's Rule.

I hope that this was helpful.

• The first thing to do is to really understand when you should use L'Hôpital's Rule.

L'Hôpital's Rule is a brilliant trick for dealing with limits of an indeterminate form.

An indeterminate form is when the limit seems to approach a deeply weird answer. For example:

""_(xrarr2) ^lim (x^2-4)/(x^2-x-2)

seems to equal $\frac{0}{0}$ if allow x to reach the value of 2.

Here's why $\frac{0}{0}$ is an impossible answer.

Let's review what the fraction $\frac{0}{5}$ means. It can be rewritten as:

$x \times 5 = 0$

and you should see that $0$ is the only possible value of x.

Now, lets look at $\frac{5}{0}$. Another way to write this fraction is:

$x \times 0 = 5$ which should seem impossible to you. Because it is.

There is no value of x that will make the above statement true. We cannot define one. So we call that answer undefined.

Now let's look at $\frac{0}{0}$. This can be rewritten as:

$x \times 0 = 0$ which is just deeply weird. It seems that x can be anything. The definition of the word limit should tell you that the answer can't be anything.

We call this an indeterminate form. There are several others, like "infinity/infinity", or "zero times infinity." None of them can possibly be the actual answer to an algebraic or logarithmic or exponential or trigonometric limit.

In the limit above, it's a bit of a trick question. I've set up what's called a removable discontinuity. We can work around the issue like so:

""_(xrarr2) ^lim (x^2-4)/(x^2-x-2)

=""_(xrarr2) ^lim ((x-2)(x+2))/((x-2)(x+1))

=""_(xrarr2) ^lim ((x+2))/((x+1)) when $x \ne 2$

which just equals $\frac{4}{3}$when we allow $x \rightarrow 2$. That's fine. That's our limit.

Now L'Hôpital's Rule says:

If you have an indeterminate form for your answer to your limit, then you can take the derivative of the numerator and of the denominator separately in order to find the limit.

You can repeat this process if you continue to get an indeterminate form.

You must stop as soon as you no longer get an indeterminate form by allowing the limit to be reached.

So for my example, we could have used L'Hôpital's Rule:

""_(xrarr2) ^lim (x^2-4)/(x^2-x-2)

""_(xrarr2) ^lim (2x)/(2x-1)

Now, since I've taken the derivative of the numerator and the denominator separately, I can try to substitute 2 in for x, and I get $\frac{4}{3}$, same as above.

If I had found the answer to still be $\frac{0}{0}$, or any other indeterminate form, then I would have to continue using L'Hôpital's Rule.

But as soon as I get a zero, or a number, or even a number over zero, I must stop.

Because when the answer is no longer an indeterminate form, L'Hôpital's Rule no longer applies.

Hope this helps.