L’ Hospitals’ Rule

L’Hôpital’s rule is a mathematical technique used to evaluate the limit of an indeterminate expression of the form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$. It states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

In other words, if $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \infty$, then $$\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f’(x)}{g’(x)}$$

provided that the derivative of the denominator is not zero.

Steps for Applying L’Hôpital’s Rule

To apply L’Hôpital’s rule, follow these steps:

1. Check if the limit of the numerator and denominator of the fraction is both 0 or both infinity.
2. If the limit of the numerator and denominator is both 0, take the derivative of the numerator and denominator.
3. Repeat step 2 until the limit of the numerator and denominator is no longer 0.
4. If the limit of the numerator and denominator is both infinity, take the derivative of the numerator and denominator.
5. Repeat step 4 until the limit of the numerator and denominator is no longer infinity.
6. Evaluate the limit of the fraction by substituting the value of $x$ that makes the limit of the numerator and denominator equal to 0 or infinity.

Examples of L’Hôpital’s Rule

Here are some examples of how to apply L’Hôpital’s rule:

Example 1: Evaluate $\lim_{x \to 0} \dfrac{\sin x}{x}$.

Solution:

1. $\lim_{x \to 0} \sin x = 0$ and $\lim_{x \to 0} x = 0$, so the limit of the numerator and denominator is both 0.
2. Taking the derivative of the numerator and denominator, we get $$\lim_{x \to 0} \dfrac{\cos x}{1} = \lim_{x \to 0} \cos x = 1.$$

Therefore, $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$.

Example 2: Evaluate $\lim_{x \to \infty} \dfrac{x^2}{e^x}$.

Solution:

1. $\lim_{x \to \infty} x^2 = \infty$ and $\lim_{x \to \infty} e^x = \infty$, so the limit of the numerator and denominator is both infinity.
2. Taking the derivative of the numerator and denominator, we get $$\lim_{x \to \infty} \dfrac{2x}{e^x} = \lim_{x \to \infty} \dfrac{2}{e^x} = 0.$$

Therefore, $\lim_{x \to \infty} \dfrac{x^2}{e^x} = 0$.

L’Hôpital’s rule is a powerful tool for evaluating the limit of indeterminate expressions. It can be used to evaluate limits of functions that have a discontinuity at a point, or limits of functions that involve transcendental functions.

Statement of L’ Hospital’s Rule

If the limit of the function $f(x)/g(x)$ as x approaches a is 0/0 or ∞/∞, and if the limit of the derivative $f’(x)/g’(x)$ as x approaches a exists, then the limit of $f(x)/g(x)$ as x approaches a is equal to the limit of $f’(x)/g’(x)$ as x approaches a.

In other words, if $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0$ or $\lim_{x \to c} f(x) = \lim_{x \to c} g(x) = \infty$, then $$\lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f’(x)}{g’(x)}$$

provided that the derivative of the denominator is not zero.

Proof of L’ Hospital’s Rule

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

If f and g are two continuous functions on the interval [a, b] and differentiable on

the interval (a, b), the $f’(c)/g(c) = [f(b)-f(a)]/[g(b)-g(a)]$ such that c belongs to $(a, b)$.

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 we have $f’(c) / g’(c)$ tends to finite limits. The functions f and g are differentiable, and $f’(x)$ and $g’(x)$ exists on the set $[c, c+k]$, and also f’ and g’ are continuous on the

interval $[c, c+k]$ provided with the conditions $f(c) = g(c) = 0$ and $g’(c) ≠ 0$ on the interval $[c, c+k]$.

By Cauchy Mean Value Theorem states that there exists CRE $(c, c+k)$, such that

$f’(c_k)/g(c_k) = [f(c+k)-f(c)]/[g(c+k)-g(c)] = f(c+k)/g(c+k)$

Now, $k→0^+$,

$ \lim_{k→0^+} \dfrac {f’(c_k)}{g’(c_k)}=\lim_{k→c^+} \dfrac {f’(x)}{g’(x)} $

while,

$ \lim_{k→0^+} \dfrac {f’(c+k)}{g’(c+k)}=\lim_{z→0^+} \dfrac {f(x)}{g(x)} $

So , we have

$ \lim_{x→c} \dfrac {f(c_k)}{g(c_k)}=\lim_{k→c} \dfrac {f’(x)}{g’(x)} $

Examples of L’ Hospital’s Rule

L’ Hospital’s rule can be used to find the limit of a variety of functions. Here are a few examples:

• Example 1: Find the limit of $(x^2 - 1)/(x - 1)$ as x approaches 1.

Solution: The limit of $(x^2 - 1)/(x - 1)$ as x approaches 1 is 0/0. So, we can use L’ Hospital’s rule to find the limit.

$$\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} = \lim_{x \to 1} \dfrac{2x}{1} = 2.$$

Therefore, the limit of $(x^2 - 1)/(x - 1)$ as x approaches 1 is 2.

$$\lim_{x \to 0} \dfrac{e^x - 1}{x} = \lim_{x \to 0} \dfrac{e^x}{1} = 1.$$

Therefore, the limit of $(e^x - 1)/x$ as x approaches 0 is 1.

Uses of L’ Hospitals’ Rule

L’Hôpital’s rule is a mathematical technique used to evaluate limits of indeterminate forms. It is particularly useful when the limit of a function is 0/0 or ∞/∞.

When to Use L’Hôpital’s Rule

L’Hôpital’s rule can be used when the limit of a function is of the form 0/0 or ∞/∞. This occurs when the numerator and denominator of the function both approach zero or infinity, respectively, as the independent variable approaches a certain value.

How to Use L’Hôpital’s Rule

To use L’Hôpital’s rule, take the derivative of the numerator and denominator of the function and then evaluate the limit of the resulting expression. If the limit of the derivative is a finite number, then that number is the limit of the original function.

Examples of Using L’Hôpital’s Rule

Here are some examples of how to use L’Hôpital’s rule to evaluate limits:

• Example 1: Find the limit of the function $f(x) = \dfrac{x^2 - 9}{x - 3}$ as $x$ approaches 3.

Solution:

The limit of the function is 0/0, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$f’(x) = \dfrac{2x}{1}$$

Evaluating the limit of the derivative, we get:

$$\lim_{x \to 3} \dfrac{2x}{1} = 6$$

Therefore, the limit of the original function is 6.

• Example 2: Find the limit of the function $g(x) = \dfrac{e^x - 1}{x}$ as $x$ approaches 0.

Solution:

The limit of the function is ∞/∞, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$g’(x) = \dfrac{e^x}{1}$$

Evaluating the limit of the derivative, we get:

$$\lim_{x \to 0} \dfrac{e^x}{1} = 1$$

Therefore, the limit of the original function is 1.

Example 3:

Find the limit of the following expression as $x$ approaches 0:

$$\lim_{x \to 0} \dfrac{\sin(x)}{x}$$

Solution:

The limit of the function is 0/0, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$\lim_{x \to 0} \dfrac{\sin(x)}{x} = \lim_{x \to 0} \dfrac{\cos(x)}{1} = 1$$

Therefore, the limit of the expression is 1.

Example 4:

Find the limit of the following expression as $x$ approaches infinity:

$$\lim_{x \to \infty} \dfrac{x^2 + 1}{x^3 - 2x + 1}$$

Solution:

The limit of the function is $ \infin / \infin $, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$\lim_{x \to \infty} \dfrac{x^2 + 1}{x^3 - 2x + 1} = \lim_{x \to \infty} \dfrac{2x}{3x^2 - 2} = 0$$

Therefore, the limit of the expression is 0.

Example 5:

Find the limit of the following expression as $x$ approaches infinity:

$$\lim_{x \to \infty} \dfrac{\ln(x)}{x}$$

Solution:

The limit of the function is $ \infin / \infin $, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$\lim_{x \to \infty} \dfrac{\ln(x)}{x} = \lim_{x \to \infty} \dfrac{1/x}{1} = 0$$

Therefore, the limit of the expression is 0.

Example 6:

Find the limit of the following expression as $x$ approaches 0:

$$\lim_{x \to 0} \dfrac{\tan(x)}{x}$$

Solution:

The limit of the function is $ 0 / 0 $, so we can use L’Hôpital’s rule. Taking the derivative of the numerator and denominator, we get:

$$\lim_{x \to 0} \dfrac{\tan(x)}{x} = \lim_{x \to 0} \dfrac{\sec^2(x)}{1} = 1$$

Therefore, the limit of the expression is 1.

L Hospital Rule FAQs

L’Hôpital’s rule is a mathematical technique used to evaluate limits of indeterminate forms. It states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, the limit of the fraction is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.

Here are some frequently asked questions about L’Hôpital’s rule:

When can I use L’Hôpital’s rule?

You can use L’Hôpital’s rule when the limit of a fraction is indeterminate, meaning it is either 0/0 or ∞/∞.

How do I apply L’Hôpital’s rule?

To apply L’Hôpital’s rule, you need to take the derivative of the numerator and denominator of the fraction and then evaluate the limit of the resulting expression.

What if the limit of the derivative is still indeterminate?

If the limit of the derivative is still indeterminate, you can apply L’Hôpital’s rule again. You can do this as many times as necessary until you reach a limit that is not indeterminate.

Are there any restrictions on using L’Hôpital’s rule?

Yes, there are some restrictions on using L’Hôpital’s rule. You cannot use L’Hôpital’s rule if the limit of the numerator or denominator is not defined. You also cannot use L’Hôpital’s rule if the derivative of the numerator or denominator is not continuous at the point in question.

What are some examples of using L’Hôpital’s rule?

Here are some examples of using L’Hôpital’s rule:

• Example 1: Find the limit of $(x^2 - 1)/(x - 1)$ as x approaches 1.

Solution:

$$lim_{(x\to1)} (x^2 - 1)/(x - 1) = lim_{(x\to1)} (2x)/(1) = 2$$

• Example 2: Find the limit of $(e^x - 1)/x$ as x approaches 0.

Solution:

$$lim_{(x\to0)} (e^x - 1)/x = lim_{(x\to0)} (e^x)/1 = 1$$

Conclusion

L’Hôpital’s rule is a powerful tool for evaluating limits of indeterminate forms. It is important to understand the restrictions on using L’Hôpital’s rule and how to apply it correctly.