Introduction to L'Hopital's rule

Calculus is a study of continuous rate of change. Sometimes this rate of change includes limits. Limit is a value to which a function approaches the output value. The L’ Hopital Rule is a rule to evaluate limits by using the first derivative. Let us learn more about L’ Hopital rule and understand how to evaluate limits by using this rule.

Understanding of the L'Hopital Rule

L'Hopital's rule is a fundamental rule of calculus that evaluates limits of indeterminate forms of functions by using derivatives. It is named after a French mathematician Guillaume-Francois-Antoine, marquis de L’ Hopital. This rule states that the limit of a quotient of two functions is equal to the limit of quotients of their derivatives.

The L Hopital’s rule is also named as Bernoulli’s rule. This rule does not directly evaluate limits but only simplifies after using cancellation tricks. In other words, we have to write the function in simpler form to evaluate the limit by using this rule.

L'Hopital's Rule Formula

The L Hopital rule states that if f(x) and g(x) are two differentiable function over an open interval I except a point c in I. Then,

• If $\lim_{x\to c} f(x) = \lim_{x→c} g(x)$ or $± ∞$, and $g’(x) ≠ 0$ for all x in I and $x ≠ c$.
• If $\lim_{x→c} \frac{f(x)}{g(x)}$ exists then
  $\lim_{x \to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$

Where $\frac{f’(x)}{g’(x)}$ is the derivative of the quotient of f(x) and g(x) which can be calculated by using the quotient rule formula of derivatives. The above formula implies that this rule allows us to approximate the derivative to a specific point where the function is not differentiable.

How to apply L'Hopital's Rule formula?

The L'Hopital's rule can be applied by finding the derivative of quotient of two functions and then taking limit to a specific point where the functions are not differentiable. But using a stepwise method to apply this rule is more suitable and accurate than just a hit and trial method. Follow these steps to apply this rule properly. 

1. Identify the function f(x) and g(x) to be differentiable at every point in an open interval I except c ≠ 0.
2. Now divide the function f(x) with g(x) and simplify the quotient if possible. 
3. Evaluate $\frac{f(x)}{g(x)} by using the limit from x to c. 
4. Find the derivative of f(x) and g(x) and divide them again. 
5. Now take the limit of the quotient of derivatives f’(x)/g’(x). 
6. If the limit of quotient of derivatives f’(x)/g’(x) is equal to the limit of quotient f(x)/g(x), the L Hopital’s rule will be verified. 

Let’s understand how the L Hopital’s rule is applied in the following examples.

L'Hopital's Rule Example 1

Find the limit of the function $\lim_{x→0} \frac{2\sin x – \sin 2x}{x – \sin x}$ by using L'Hopital's rule. 

Given function is, 

$\lim_{x→0} \frac{2\sin x – \sin 2x}{x – \sin x}$

If we directly substitute the limit, the function will be indeterminate. 

Suppose that 

$f(x) = 2\sin x – \sin 2x$

$g(x) = 2 – \sin x$

Now differentiating both one by one, we get,

$f’(x) = 2\cos x – 2\cos 2x$

$g’(x) = 1 – \cos x$

Where the derivative of sin x is cos x.

Now simplify the quotient of derivatives f’(x) and g’(x)

$\lim_{x\to 0}\frac{f'(x)}{g'(x)}=\lim_{x\to 0}\frac{2\cos x-2\cos 2x}{1-\cos x}$

$\lim_{x\to 0}\frac{f'(x)}{g'(x)}= \lim_{x\to 0}(4\cos x +2)$

Substituting the limit, 

$\lim_{x\to 0}\frac{f'(x)}{g'(x)} = 4\cos 0+2 = 6$

L'Hopital's Rule example 2

Evaluate the limit $\lim_{x→2}\frac{x^2 + x+6}{x^2 – 4}$ by using L'Hopital's rule.

Given that

$\lim_{x→2}\frac{f(x)}{g(x)}=\lim_{x→2}\frac{x^2 + x+6}{x^2 – 4}$

Where, 

$f(x) = x^2 + x+6$

$g(x) = x^2 – 4$

f(x) and g(x) both contain polynomial functions, therefore we can use the power rule of the derivative.

Differentiating f(x) and g(x) with respect to x.

$f’(x) = 2x +1$

And 

$g’(x) = 2x$

Dividing derivatives of f(x) and g(x)

$\lim_{x→2}\frac{f’(x)}{g’(x)} = \lim_{x→2}\frac{2x + 1}{2x}$

Now substituting the limit x approaches to 2, 

$\lim_{x\to 2}\frac{f'(x)}{g'(x)} =\frac{2(2) + 1}{2(2)}$

$\lim_{x→2}\frac{f’(x)}{g’(x)} = \frac{5}{4}$

L'Hopital's Rule example 3

Evaluate the limit by using L'Hopital's rule $\lim_{x→0}\frac{x}{e^{2x} – 1}$.

Given that, 

$\lim_{x→0}\frac{x}{e^{2x} – 1}$

Suppose that, 

$f(x) = x$

$g(x) = e^{2x} – 1$

Differentiating with respect to x. 

$f’(x) =1$

$g’(x) = 2e^{2x}$

Where, the derivative of x is 1.

Dividing derivatives of f(x) and g(x)

$\lim_{x→0}\frac{f'(x)}{g'(x)} = \lim_{x\to 0} \frac{1}{2e^{2x}}$

Now substituting the limit x approaches to 0, 

$\lim_{x→0}\frac{f'(x)}{g'(x)} = \frac{1}{2(e^0)}$

Since $e^0 = 1$

$\lim_{x→0}\frac{f'(x)}{g'(x)} =\frac{1}{2}$

Comparison between L'Hopital's and Quotient Rule

The comparison between the L'Hopital's rule and quotient rule can be easily analysed using the following difference table.

Conclusion

The L Hopital’s rule is a fundamental rule of derivatives which includes limits. If a function is differentiable over an open interval I except c a point in I, the L'Hopital's rule is used to approximate that function. It is useful to solve limits where the direct substitution of limits makes 0/0 form. e.g., we obtain 0 in denominator as well as in numerator.