L Hospital Rule

This is a rule that helps to evaluate the limits which involve indeterminate forms by using the derivatives. An indeterminate form can be defined as a limit that does not provide enough information to determine the original limit. It is a very important rule in Calculus. With this rule, we can actually find the value of certain kinds of limits using derivatives. This rule is named after a person. In 1696, a French mathematician named Guillaume François Marquis De L'Hospital, where “L’Hospital” is pronounced as “low-pee-tal” and not “le-hoss-pih-tal”.

What is L'Hospital’s Rule?

While solving sums on limits, it is possible to face a deadlock when the numerator and the denominator of the limit gives a 0/0 or \[\infty\] / \[\infty\] situation. This is called an indeterminate form. In such a case, to move ahead and compute the limit, by differentiating both the numerator and the denominator till the numerator and the denominator no longer give the indeterminate form is a way out. This process of taking derivatives of the numerator and the denominator of the limit is known as the L'Hospital rule.

This rule can be applied more than once as well. Even if we apply this rule once it still holds an indefinite form every time after its applications. But if the problem is out of the indeterminate forms, then l hospital rule cannot be applied. 

L Hospital Rule Formula

If f(x)/g(x) is in the form 0/0 or \[\infty\] / \[\infty\] when x=a plugs in, then:      

\[\lim_{x\rightarrow a}\] \[\frac{f(x)}{g(x)}\] = \[\lim_{x\rightarrow a}\] \[\frac{{f}'(x)}{{g}'(x)}\]

Basically we just have to take the derivative of the numerator as well as of the denominator and compute the limit. 

Proof of L'Hospital's Rule

Using the Extended Mean Value Theorem or Cauchy’s Mean Value Theorem, L’Hospital’s rule can be proved.

If f and g are two continuous functions on the interval (a, b) and are differentiable on the interval (a, b), then

\[\frac{f’(c)}{g’(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}\], such that c  ∈  (a, b)

Assume that the two functions f and g are defined on the interval (c, b) in such a way that f(x)→0 and g(x)→0, as x→c+

But, f’(c) / g’(c) tends to finite limits. The functions f and g are differentiable, and f’(x) and g’(x) exists on the set (c, c+k)

Also f’(x) and g’(x) are continuous on the interval (c, c+k) given that f(c)= g(c) = 0 and g’(c) ≠ 0 on the interval (c, c+k)

By Cauchy’s Mean Value Theorem, there exists ck∈ (c, c+k), such that

\[ \frac{f’(c_{k})}{g’(c_{k})}\] = \[\frac{f(c+k)-f(c)}{g(c+k)-g(c)}\] = \[\frac{f(c+k)}{g(c+k)}\]

Now, k→0+,

\[ \lim_{k\rightarrow 0^{+}} \] \[ \frac {f’(c_{k})}{g’(c_{k})}\] = \[ lim_{x\rightarrow c^{+}} \] \[ \frac {f’(x)}{g’(x)}\]

Also, \[ \lim _{k\rightarrow 0^{+}}\] \[\frac{f(c+k)}{g(c+k)}\] = \[\lim_{x\rightarrow 0^{+}} \frac{f(x)}{g(x)}\]

Then, we can have

\[ \lim_{x\rightarrow c^{+}}\] \[\frac{f(x)}{g(x)}\] = \[ \lim _{x\rightarrow c^{+}}\] \[\frac{f’(x)}{g’(x)} \]

L Hospital Rule

If we desire to draw advantage from this rule then we also have to check that the limit is in the right form. And that is done in the following way:

1. To apply this rule we have to make sure that the fraction must be of two functions, that is f(x)/g(x)

2. It is very crucial to see that when you plug in the x-value, the function must evaluate to either 0/0 or \[\infty\] / \[\infty\] as these are the two types of indeterminate forms. We won’t be able to use this method directly if the limit problem is not in an indeterminate form. 

Uses of L’Hospital’s Rule

Using L'Hospital's rule, you will be able to solve the problem in 0/0,\[\infty\] / \[\infty\],\[\infty\] - \[\infty\], 0.\[\infty\], \[1^{\infty}\], \[\infty^{0}\], or \[0^{0}\]forms. These forms are known as indeterminate forms. To remove the indeterminate forms in the problem, we can solve a fraction under a certain limit by using L’Hospital’s rule.

L Hospital Rule Example

Here are few l hospital rule problems with solutions. 

Example 1) \[\lim_{x\rightarrow 0}\]  \[\frac{sin(4x)}{7x - 2x^{2}}\]

Solution 1) Now by plugging in x = 0, we will find the indeterminate form, 0/0. And thus, L’Hospital’s can be used by taking the derivative of the top and the bottom. Then we can plug in the value x. 

\[\lim_{x\rightarrow 0}\] \[\frac{sin(4x)}{7x - 2x^{2}}\] = \[\lim_{x\rightarrow 0}\] \[\frac{4cos(4x)}{7-4x}\] = \[\frac{4cos(4(0))}{7-4(0)}\] \[\frac{4}{7}\]

Example 2) \[\lim_{x\rightarrow \infty}\] \[\frac{3x^{2} - 2x + 1}{5x^{2} + 17}\]

Solution 2) This time the indeterminate for is  \[\infty\] / \[\infty\]

There is one interesting thing about this example and that is even after using it once, the limit still has the same indeterminate form and therefore we can use it once more. 

\[\lim_{x\rightarrow \infty}\]  \[\frac{3x^{2} - 2x + 1}{5x^{2} + 17}\] = \[\lim_{x\rightarrow \infty}\] \[\frac{6x-2}{10x}\] = \[\lim_{x\rightarrow \infty}\] \[\frac{6}{10}\] = \[\frac{3}{5}\]

Basically, we can use the L’Hospital rule as many times as possible with the only condition that there has to be an indeterminate form at each and every stage.