L'Hospital's Rule 

L’Hospital’s Rule, also known as L'Hopital's Rule, is a powerful tool in calculus for finding limits when you get indeterminate forms like _\frac00   or  _{\frac{\infty}{\infty}} . When direct evaluation gives confusing results, L'Hopital’s Rule allows you to take the derivative of the numerator and the denominator, then find the limit again. This makes solving tricky limits much easier. It’s a key concept in calculus that simplifies many limit problems, helping you tackle them more effectively and quickly.

1.0What is L'Hopital’s Rule?

L'Hopital's Rule is a mathematical method used to find limits of indeterminate forms. An indeterminate form occurs when you try to evaluate a limit and get an expression like $\frac{0}{0}$ or  $\frac{\infty }{\infty }$ . These forms don’t give a clear answer, so L'Hopital’s Rule helps us simplify the problem and find the correct limit.

The rule states that if you have a limit of the form:

${\lim }_{x\to c}\frac{f\left(x\right)}{g\left(x\right)}$

where both f(x) and g(x) approach 0 or ∞ as  $x\to c$, you can apply the following formula:

${\lim }_{x\to c}\frac{f\left(x\right)}{g\left(x\right)}={\lim }_{x\to c}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

This means that instead of directly evaluating the limit, you can take the derivative of the numerator (f'(x)) and the derivative of the denominator (g'(x)), and then find the limit of the new fraction.

2.0When to Use L'Hopital's Rule

L'Hopital's Rule is used when you encounter indeterminate forms like:

1. 0/0 – Both the numerator and denominator approach zero.
2. ∞/∞ – Both the numerator and denominator approach infinity.

These situations often arise in calculus when dealing with limits of rational functions. If you encounter these forms, don’t panic! L'Hopital’s Rule can save you time and effort.

3.0Solved Example on L’Hôpital’s Rule 

Example 1: Find the limit: ${\lim }_{x\to 0}\frac{\sin \left(x\right)}{x}$

At first glance, both the numerator and denominator approach 0 as $x\to 0$ , which is an indeterminate form (0/0). Applying L'Hopital's Rule:

1. Take the derivative of the numerator: $\frac{d}{\mathrm{d}x}\left[\sin \left(x\right)\right]=\cos \left(x\right)$
2. Take the derivative of the denominator: $\frac{d}{\mathrm{d}x}\left[x\right]=1$

Now the limit becomes:

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

$=\frac{\cos \left(0\right)}{1}$

$=\frac{1}{1}=1$

So, the limit is 1.

Example 2: ${\lim }_{x\to 0}\frac{x-\sin x}{x^3}$

Solution:

This is of the form 0/0. Apply L'Hopital's Rule.

Using Formula

${\lim }_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}={\lim }_{x\to c}\frac{f^{\prime }\left(x\right)}{g^{\prime }\left(x\right)}$

Numerator: $1-\cos x$

Denominator: $3x^2$

Now, take the limit:

${\lim }_{x\to 0}\frac{1-\cos x}{3x^2}$

This is still 0/0, so we apply L'Hopital's Rule again.

Differentiate again:

Numerator:

Denominator: 6x

Now, take the limit:

${\lim }_{x\to 0}\frac{\sin x}{6x}=\frac{0}{0}\left(Againapply{L}^{\prime }Hopita{l}^{\prime }sRule\right)$

Differentiate again:

Numerator: $\cos x$

Denominator: 6

Now, take the limit: ${\lim }_{x\to 0}\frac{\cos x}{6}=\frac{1}{6}$

Thus, the limit is 1/6.

Example 3: ${\lim }_{x\to 0}\frac{x-\sin x\cos x}{x^2}$

Solution:

This is of the form 0/0. Apply L'Hopital's Rule.

Differentiate the numerator and denominator:

Numerator: 

$\cos x\cos x-\sin x\left(-\sin x\right)$

$=2{\cos }^{2}x+{\sin }^{2}x$

Denominator: 2x

Now, take the limit:

${\lim }_{x\to 0}\frac{2{\cos }^{2}x+{\sin }^{2}x}{2x}=\frac{2\left(1\right)+0}{0}=Indeterminateform$. So, apply L’Hopital’s rule again.

Example 4: ${\lim }_{x\to 0}\frac{\tan 2x-\tan x}{x}$

Solution:

This is of the form 0/0, so we apply L'Hopital's Rule.

Differentiate the numerator and denominator:

Numerator: $2{\sec }^2\left(2x\right)-{\sec }^2\left(x\right)$

Denominator: 1

Now, take the limit:

${\lim }_{x\to 0}\frac{2{\sec }^2\left(2x\right)-{\sec }^2\left(x\right)}{1}$

$=2{\sec }^2\left(0\right)-{\sec }^2\left(0\right)$

$=2-1=1$

Thus, the limit is 1.

Example 5: ${\lim }_{x\to 0}\frac{1-\cos 3x}{x}$

Solution:

This is of the form 0/0. Apply L'Hopital's Rule.

Differentiate the numerator and denominator:

Numerator: $3\sin \left(3x\right)$

Denominator: 1

Now, take the limit: ${\lim }_{x\to 0}\frac{3\sin \left(3x\right)}{1}=3\left(0\right)$

Thus, the limit is 0.

Example 6: ${\lim }_{x\to 0}\frac{\sin 2x-2x}{x^3}=3\left(0\right)$

Solution:

This is of the form 0/0. Apply L'Hopital's Rule.

Differentiate the numerator and denominator:

Numerator: $2\cos \left(2x\right)-2$

Denominator: $3x^2$

Now, take the limit:

${\lim }_{x\to 0}\frac{2\cos \left(2x\right)-2}{3x^2}={\lim }_{x\to \infty }\frac{2\left(1\right)-2}{0}=Indeterminate$. Apply L’Hopital’s Rule again.