Subject: Calculus

# L Hôpitals Rule

In our previous article regarding limits, we were thought of different techniques, theorems (aka "shortcut rules"), and examples in evaluating the limit of a function. In fact, we were provided with many rules and theorems that get us hard time memorizing or at least retaining it in our memory. But those techniques are not as easy as it may get when it comes to limits with indeterminate forms. For example, the limit

\lim_{x \to 1}\frac{x^3-1}{x^2-1}

if evaluated directly would give an indeterminate form 0\over 0. But thanks to the new technique that could totally help us in these kinds of limits and this technique is more easier and fun to learn. This technique is called L Hôpitals Rule which is discovered by a 17th-century French mathematician Guillaume de l'Hôpital, hence the name.

## Why this time? Why not present that theorem directly before?

Well, unfortunately this theorem uses differentiation and to know that differentiation is not yet introduced the time we were discussing limits, then it is best to keep this as a secret until we already discussed about derivative.
But before that, it must be kept in mind that only limits with indeterminate foms such as \frac{0}{0} and \infty \over \infty applies to this technique. Other than that, then L Hôpitals Rule can never be applied.

## Okay, I'm excited. Get me directly into it.

Let me introduce to you the following theorem on L Hôpitals Rule first then we'll go from there.

### Theorem (L Hôpitals Rule)

 Supose that the functions f and g are differentiable functions and suppose further that a is a finite number or \infty. Then if we have \lim_{x \to a}f(x)=0 and \lim_{x \to a}g(x)=0, then the limit \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}=\lim_{x \to a}f'(x)\over \lim_{x \to a}g'(x).

The same goes for indeterminate form \infty \over \infty,

 Supose that the functions f and g are differentiable functions and suppose further that a is a finite number or \infty. Then if we have \lim_{x \to a}f(x)=\infty and \lim_{x \to a}g(x)=\infty, then the limit \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}=\lim_{x \to a}f'(x)\over \lim_{x \to a}g'(x).

Recall that \frac{\lim_{x \to a}f(x)}{\lim_{x \to a}g(x)}=\lim_{x \to a}\frac{f(x)}{g(x)}, thus, we can easily extend L Hôpitals Rule to evaluating \lim_{x \to a}\frac{f(x)}{g(x)} if and only if the above conditions are met. So that's it. L Hôpitals Rule makes it easy for us to evaluate indeterminate limits by simply taking the limit of the derivatives of that functions (see example #1). Also, if after taking the first derivative of both upper and lower function and the result is still indeterminate, we will again apply the same rule up until it is no longer indeterminate (see example #2).

### Example #1

Let us evaluate the following limits by using L Hôpitals Rule.

\lim_{x \to 1}\frac{x^3-1}{x^2-1}

We know that direct evaluation will result to an answer 0 \over 0. This is an indeterminate form, thus we'll use the rule presented above. Accordingly, we will differentiate the numerator and the denominator separately, thus,

f'(x)=D_{x}(x^3-1)=3x^2

also,

g'(x)=D_{x}(x^2-1)=2x, thus,

\lim_{x \to 1}\frac{x^3-1}{x^2-1}=\lim_{x \to 1}\frac{3x^2}{2x}=\lim_{x \to 1}\frac{3x}{2}

Thus, the answer is 3\over 2.

### Example #2

Evaluate the following limits by L Hôpitals Rule.

\lim_{x \to 0}\frac{1-\cos{x}}{x^2}

Again, direct evaluation will result to an answer 0 \over 0. This is an indeterminate form, thus we'll use the rule presented above. Accordingly, we will differentiate the numerator and the denominator separately, thus,

f'(x)=D_{x}(1-\cos{x})=\sin{x}

also,

g'(x)=D_{x}(x^2)=2x, thus,

\lim_{x \to 0}\frac{x^3-1}{x^2-1}=\lim_{x \to 0}\frac{\sin{x}}{2x}=\frac{0}{0}

This answer is still an indeterminate form, thus, we apply again L Hôpitals Rule which will then be,

f''(x)=D_{x}(1-\cos{x})=\cos{x}

also,

g''(x)=D_{x}(x^2)=2, thus,

\lim_{x \to 0}\frac{x^3-1}{x^2-1}=\lim_{x \to 0}\frac{\cos{x}}{2}=\frac{1}{2}

Thus, by applying L Hôpitals Rule twice, we finally get the finite answer to the problem which is 1 \over 2.

### Example #3

Solve the following limit.

\lim_{x \to 0^+}x^x

ANSWER: \lim_{x \to 0^+}x^x=1

### Example #4

Find the limit of the following.

\lim_{z \to 0}\frac{\sin{z}}{z}

ANSWER: \lim_{z \to 0}\frac{\sin{z}}{z} = 1

### Example #5

Solve the following limit.

\lim_{x \to 0}\frac{\sin{x}-\tan{x}}{x^3}

ANSWER: \lim_{x \to 0}\frac{\sin{x}-\tan{x}}{x^3}=-\frac{1}{2}

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