If you think that we are going to give you an experience of how a hospital runs, then don’t worry because we are not trodding on a medical field. 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”.

L Hospital rule is a method that helps to evaluate indeterminate forms such as 0/0 or ∞/∞. In order to evaluate the limits of indeterminate forms for the derivatives in calculus, we use L'Hospital's. 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.

If f(x)/g(x) is in the form 0/0 or ∞/∞ 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.

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:

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

It is very crucial to see that when you plug in the x-value, the function must evaluate to either 0/0 or ∞/∞ 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.

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.

FAQ (Frequently Asked Questions)

1. When can L’hospital method be not applicable?

L’Hospital rule must satisfy 4 major conditions or else it would be proved as not applicable. These 4 major conditions are:

f and g must be differentiable in an open interval.

The well-known condition for L’Hospital is that limit must be of the form 0/0 or ∞/∞

The third restriction, g’(x) 0 is very necessary for the general proof of L’hopital’s rule which is demonstrated beautifully by the counterexample provided by Austrian mathematician Otto Stolz.

At last, the condition that the limit x → af'(x)/g'(x)should exist as demonstrated by the counterexample lim x → (x+sinx)/x

2. Can we prove L’Hospital graphically?

let us consider x = c and imagine that both the functions f(x) and g(x) are expanding about c. then:

f (x) ≈ f (c) + f' (c) (x-c)

g (x) ≈ g (c) + g' (c) (x-c)

Now since f(c) = g(c) = 0 when we get too close to c, we have:

f(x) ≈ f' (c) (x-c)

g (x) ≈g' (c) (x-c)

If we graph it geometrically, there will be two straight lines with two different non-zero slopes and both the lines cross the x axis at x=c.

If we try to drive it algebraically, we are left to first order with the ratio of the derivatives. This illustrates the geometrical reality that to first order, the ratio of the two functions as you get close to c is just the ratio between the slopes of the two lines.