### L'Hospital Rule

Guillaume de l'HÃ´pital, famously known as L'Hospital, came up with a method to deal with fractions, when they take the **indeterminate** forms, approaching certain limits.

The indeterminate form can be as follows:

- 0/0 - as in the case of
**sin(x)/x**, when x approaches 0 - ∞ / ∞ - as in the case of
**e**, when x approaches ∞^{x2}/ x^{2}

L'Hospital Rule help us deal with situations of this kind. It is as follows:

#### L'Hospital Rule

If Limit[f(x)/g(x)] as x approaches a is 0/0 or ∞ / ∞, then Limit[f(x)/g(x)] as x approaches a is [f'(x)/g'(x)].In all the following animations, **[f(x)/g(x)]** is drawn in **red** and **[f'(x)/g'(x)]** in **purple.**

Please note the **convergence** pf the two curves / lines to the same point, as the limit approaches.

**E.g.1**

Find Limit [(4x-3)/(5x-6)] as x approaches ∞.

If x = ∞, then [(4x-3)/(5x-6)] = ∞/∞ - indeterminate

Let's use L'Hospital rule for this:

f'(x)/g'(x) = 4/5 = 0.8, as x approaches ∞

**E.g.2**

Find Limit [(x-4)/ln(x-3)] as x approaches 4.

If x = 4;, then [(x-4)/ln(x-3)] = 0/0 - indeterminate

Let's use L'Hospital rule for this:

f'(x)/g'(x) = 1/(1/x-3) = 1, as x approaches 4.

**E.g.3**

Find Limit [ln(x)/√x] as x approaches ∞.

If x = ∞, then [ln(x)/√x] = ∞/∞- indeterminate

Let's use L'Hospital rule for this:

f'(x)/g'(x) = (1/x)/(1/2)x^{-1/2} = 2/√x = 0, as x approaches ∞.

The behaviour of the curve will be clearer when x is really large.

**E.g.4**

Find Limit [(x^{2} -x - 6)/(x^{2} -3x)] as x approaches 3;.

If x = 3, then [(x^{2} -x - 6)/(x^{2} -3x)] = 0/0 - indeterminate

Let's use L'Hospital rule for this:

f'(x)/g'(x) = (2x-1)/(2x-3) = 5/3, as x approaches 3.

**E.g.5**

Find Limit sin(x) /x as x approaches 0 and hence sketch y = sin(x)/x.

sin(x) /x when x approaches 0 = sin(0)/0 = 0/0 - indeterminate

Let's use L'Hospital rule for this:

f'(x) = cos(x); g'(x) = 1

So, f'(x)/g'(x) = cos(x)/1

When x approaches 0, f'(x)/g'(x) = 1/1 = 1

Therefore, sin(x)/x, when x approaches 0 = 1.