# L'Hopital's Rule: Evaluating Limits of Indeterminate Forms

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Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.

Do you know how to solve for the value of the limit of the function below?

limx→1 ln(x)/x−1

In computing this limit, we can't apply quotient limit law because the denominator's limit is 0. Although the function's limit exists, its value is not evident because both numerator and denominator approach 0 and 0/0 is not defined.

In general, if we have a limit of the form limx→a f(x)/g(x) where both f(x)→0 and g(x)→0 as x→a, then this limit may or may not exist and is called an indeterminate form of type 0/0.

## Indeterminate Forms of Type ∞/∞

Another situation in which a limit is not apparent occurs when we look for a horizontal asymptote of the function and need to evaluate the limit.

limx→∞ ln(x)/x−1

It isn't obvious how to evaluate this limit because both numerator and denominator become large as x→∞. There is a struggle between numerator and denominator. If the numerator wins, the limit will be ∞; if the denominator wins, the answer will be 0. Or there may be some compromise, in which case the answer will be some finite positive number.

In general, if we have a limit of the form limx→a f(x)/g(x) where both f(x)→∞ (or−∞) and g(x)→∞ (or−∞) as x→a, then this limit may or may not exist and is called an indeterminate form of type ∞/∞.

## What Is L'Hopital's Rule?

L'Hopital's Rule says that the limit of a quotient of functions is equal to the limit of their derivatives' quotient, provided that the given conditions are satisfied. Suppose f and g are differentiable and g′(x) ≠ 0 near a (except possibly at a). Suppose that

limx→a f(x)/g(x)

Where:

1. limx→a f(x) = 0 and limx→a g(x) = 0
2. limx→a f(x) = ±∞ and limx→a g(x) = ±∞

In other words, we have an indeterminate form of type 0/0 or ∞/∞
Then

limx→a f(x)/g(x) = limx→a f′(x)/g′(x)

If the limit on the right side exists (or is ∞ or −∞)

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## L’Hopital’s Rule Proof

L'Hopital's Rule is named after a French nobleman, the Marquis de l'Hopital (1661-1704), but initially discovered by a Swiss mathematician, John Bernoulli (1667-1748). You might sometimes see L'Hopital spelt as L'Hospital, as was common in the 17th century.

The figure below suggests visually why L'Hopital's Rule might be valid. The first graph shows two different functions f and g, each which approaches 0 as x→a.

If we were to zoom in toward the point (a, 0), the graphs would start to look almost linear. But if the functions were linear, as in the second graph, then their ratio would be

m1(x−a)/m2(x−a) = m1/m2

As you can observe, this is the ratio of their derivatives. It suggests that

limx→a f(x)/g(x) = limx→a f′(x)/g′(x)

The available version of L'Hopital's Rule proof is a difficult one to prove. You can find a more detailed explanation in advanced textbooks or references.

## When to Use L'Hopital's Rule

It is essential to verify the conditions regarding the limits of variables f and g before using L'Hopital's Rule.

• You may use the L'Hopital's Rule for limits with an indeterminate form c or ∞/∞, as described earlier. You have to differentiate the numerator and the denominator before taking the limits.
• L'Hopital's Rule is also valid for one-sided limits and limits at infinity or negative infinity; that is, "x→a" can be replaced by any of the symbols x→a+, x→a−, x→∞, or x→−∞.
• For the special case in which f(a) = g(a) = 0, f' and g' are continuous, and g′(a) ≠ 0, it is easy to see why l'Hopital's Rule is true. Using the alternative form of the definition of a derivative, we have

limx→a f′(x)/g′(x) = limx→a f′(a)/g′(a)

limx→a f′(x)/g′(x) = [limx→a (f(x) − f(a)/x-a)] / [limx→a (g(x) − g(a)/x−a)]

limx→a f′(x)/g′(x) = limx→a (f(x) − f(a)x−a(x)−g(a)x−a)

limx→a f′(x)/g′(x) = limx→a f(x) − f(a)/g(x)−g(a) = limx→a f(x)g(x)

limx→a f′(x)/g′(x) = limx→a f(x) − f(a)/g(x)−g(a)

limx→a f′(x)/g′(x) = limx→a f(x)/g(x)

There are more other indeterminate forms. Some of these are indeterminate products, indeterminate differences, and indeterminate powers.

## Example 1: An Indeterminate Form of Type 0/0

Evaluate the limit of limx1 (ln(x)) / (x−1) using the L’Hopital’s Rule.

Solution

Solve the limit of the numerator and the denominator separately.

limx1 ln(x) = ln(1) = 0

limx1 (x − 1) = 0

Apply L’Hopital’s Rule. Differentiate the numerator and denominator separately and do not use the Quotient Rule.

limx1 ln(x) / x−1 = limx1([d/dx](lnx) / [d/dx](x−1))

limx1 1/x = 1

The limit of the ln(x) / (x−1) as x approaches one is 1.

## Example 2: An Indeterminate Form of Type ∞/∞

Calculate limx→ [ex] / [x2] by applying L’Hopital’s Rule.

Solution

We have limx→ ex = ∞ and limx→ x2 = ∞, so L’Hopital’s Rule gives

limx→ [ex] / [x2] = limx→ (d/dx) [ex] / (d/dx) [x2]

limx→ [ex] / [x2] = limx→ [ex] / [2x]

Since ex → ∞ and 2x → ∞ as x approaches ∞, the limit on the right side is also indeterminate, but the second application of L’Hopital’s Rule gives

limx→ [ex] / [x2] = limx→ [ex] / [2x]

limx→ [ex] / [x2] = limx→ [ex] / [2]

limx→ [ex] / [x2] = ∞

The limit of limx→ [ex] / [x2] is ∞.

## Example 3: An Indeterminate Form of Type ∞/∞

Evaluate the limit of the function limx→ [ln(x)] / [3√x].

Solution

Since ln(x) → ∞ and 3√x → ∞ as x → ∞, L’Hopital’s Rule applies:

limx→ [ln(x)] / [3√x] = limx→ [1/x] / [(1/3)x-2/3]

Notice that the limit on the right side is now indeterminate of type 0/0. But instead of applying L’Hopital’s Rule a second time, we simplify the expression and see that a second application is unnecessary.

limx→ [ln(x)] / [3√x] = limx→ [1/x] / [(1/3)x-2/3]

limx→ [ln(x)] / [3√x] = limx→[3] / [3√x]

limx→ [ln(x)] / [3√x] = 0

The limit of the function limx→ [ln(x)] / [3√x] is zero.

## Example 4: Using the L’Hopital’s Rule Three Times

Use the L’Hopital’s Rule and find the limit of the function [tan(x) – x] / x3 as x approaches 0.

Solution

Noting that both [tan(x) – x] → 0 and x3 → 0 as x → 0, we use L’Hopital’s Rule:

limx→0 [tan(x) – x] / x3 = limx→0 [sec2 (x) – 1] / 3x2

Since the limit on the right side is still indeterminate of type 0/0, we apply L’Hopital’s Rule again:

limx→0 [sec2 (x) – 1] / 3x2 = limx→0 [2sec2(x) tan(x)] / 6x

Because limx→0 sec2(x) = 1, we simplify the calculation.

limx→0 [2sec2(x) tan(x)] / 6x = (1/3) [limx→0 sec2(x)] [limx→0 tan(x) / x]

limx→0 [2sec2(x) tan(x)] / 6x = (1/3) [limx→0 tan(x) / x]

We can evaluate this last limit either by using L’Hopital’s Rule a third time or by writing tan(x) as sin(x)/cos(x) and making use of our knowledge of trigonometric limits. Putting together all the steps, we get the following.

limx→0 [tan(x) – x] / x3 = limx→0 [sec2 (x) – 1] / 3x2

limx→0 [tan(x) – x] / x3 = limx→0 [2sec2(x) tan(x)] / 6x
limx→0 [tan(x) – x] / x3 = (1/3) limx→0 [tan(x)] / x

limx→0 [tan(x) – x] / x3 = (1/3) limx→0 [sec2(x)] / 1

limx→0 [tan(x) – x] / x3 = 1/3

The limit of the function [tan(x) – x] / x3 as x approaches zero is 1/3.

## Example 5: Finding the Limit of a Function with Sine and Cosine

Find the limit of limx→π- [sin(x)] / [1 – cos(x)].

Solution

If we blindly attempted to use L’Hopital’s Rule, we would get the following.

limx→π- [sin(x)] / [1 – cos(x)] = limx→π- [cos(x)] / [sin(x)]

limx→π- [sin(x)] / [1 – cos(x)] = -∞

This is wrong! Although the numerator sin(x) → 0 as x → π, notice that the denominator 1 – cos(x) does not approach 0, so L’Hopital’s Rule can’t be applied here. The required limit is, in fact, easy to find because the function is continuous at π and the denominator is nonzero there:

limx→π- [sin(x)] / [1 – cos(x)] = limx→π- [sin(π)] / [1 – cos(π)]

limx→π- [sin(x)] / [1 – cos(x)] = 0 / [1 – (-1)]

limx→π- [sin(x)] / [1 – cos(x)] = 0

It shows that you can go wrong if you use L’Hopital’s Rule without thinking. So when evaluating any limit, you should consider other methods before using L’Hopital’s Rule.

The limit of sin(x) / [1 – cos(x)] as x → π is 0.

## Example 6: Evaluating the Limit of a Sine Equation in 0/0 Indeterminate Form

Evaluate the limit limx→0 sin(x) / x.

Solution

In the given equation, both the numerator and denominator have limits 0. It implies that the equation is a 0/0 indeterminate form which means we need to apply the L’Hopital's Rule.

limx→0 [sin (x)] / x = [sin (0)] / 0 = 0/0

Apply the L’Hopital's Rule by differentiating the numerator and denominator separately.

limx→0 [sin (x)] / x = limx→0 [cos (x)] / 1 = 1/1 = 1

Therefore, the limit of sin(x) / x as x approaches zero is 1.

## Example 7: Finding the Limit of an Equation with 0/0 Indeterminate Form

Evaluate the limit of limx→3 [x-3] / [x2 – 9].

Solution

Substituting x = 3 to the equation [x-3] / [x2 – 9] results to a 0/0 indeterminate form. Apply the L’Hopital's Rule and differentiate the top and bottom separately.

limx→3 [x-3] / [x2 – 9] = limx→3 [1 – 0] / [2x – 0]

limx→3 [x-3] / [x2 – 9] = limx→3 1/2x

Substitute the value x = 3 to the derivative of the equation.

limx→3 [x-3] / [x2 – 9] = 1 / 2(3)

limx→3 [x-3] / [x2 – 9] = 1 / 6

The limit of [x-3] / [x2 – 9] as x approaches three is 1/6.

## Example 8: Finding the Limit of an Euler Equation

Evaluate the limit of limx→∞ [e4x] / [10x +500].

Solution

The Euler equation given is in a ∞/∞ indeterminate form.

limx→∞ [e4x] / [10x +500] = limx→∞ [e(∞)(4)] / [10(∞) +500]

limx→∞ [e4x] / [10x +500] = limx→∞ [e] / ∞

Apply the L’Hopital's Rule. Differentiate the numerator using the chain rule.

limx→∞ [e4x] / [10x +500] = limx→∞ [4e4x] / [10 + 0]

limx→∞ [e4x] / [10x +500] = [4e] / [10]

limx→∞ [e4x] / [10x +500] = ∞ / 10

limx→∞ [e4x] / [10x +500] = ∞

The limit of [e4x] / [10x +500] as x approaches ∞ is ∞.

## Example 9: Finding the Limit of a Complex Quadratic Equation

Find the limit of limx→1 [x2 – 1] / [x2 + 3x – 4].

Solution

Substituting x = 1 to the equation [x2 – 1] / [x2 + 3x – 4] results to a 0/0 indeterminate form. Apply the L’Hopital's Rule and differentiate the top and bottom separately.

limx→1 [x2 – 1] / [x2 + 3x – 4] = limx→1 [12 – 1] / [x2 + 3x – 4] = 0/0

Get the derivative of the numerator and denominator separately.

limx→1 [x2 – 1] / [x2 + 3x – 4] = limx→1 [2x – 0] / [2x + 3 – 0]

limx→1 [x2 – 1] / [x2 + 3x – 4] = limx→1 [2x] / [2x + 3]

limx→1 [x2 – 1] / [x2 + 3x – 4] = limx→1 [2(1)] / [2(1) + 3]

limx→1 [x2 – 1] / [x2 + 3x – 4] = 2 / 5

The limit of [x2 – 1] / [x2 + 3x – 4] as x approaches one is 2/5.

## Example 10: Finding the Limit of a Tangent and Sine Equation

Evaluate the limit of limx→0 [sin (x2)] / [x tan (x)].

Solution

The given equation is of the form 0/0 indeterminate form.

limx→0 [sin (x2)] / [x tan (x)] = limx→0 [sin (02)] / [0 tan (0)] = 0/0

Apply the L’Hopital's Rule. Obtain the derivative of the numerator and denominator separately. Apply the Chain Rule in Calculus on the numerator containing the sine function and use the Product Rule on the denominator having the tangent function.

limx→0 [sin (x2)] / [x tan (x)] = limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)]

limx→0 [sin (x2)] / [x tan (x)] = limx→0 [2(0) cos(0)2] / [(0) sec2 (0) + (1) tan(0)]

limx→0 [sin (x2)] / [x tan (x)] = limx→0 [2(0)(1)2] / [(0) (1)2 + (0)]

limx→0 [sin (x2)] / [x tan (x)] = limx→0 [0] / [(0) (1)2 + (0)] = 0 / 0

Next, apply the L’Hopital's for the second time. Now, use both the Product Rule and the Chain Rule for the numerator and the denominator.

limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)] = [(2x) (2x) (-sin x2) + (2) cos (x2)] / [(x) (2 sec(x)) (sec (x) tan (x) + (1) sec2 (x) + sec2(x)]

limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)] = [(2) (-cos x2) – (4x2) sin (x2)] / [(2x) (sec2(x)) tan (x) + (2) sec2 (x)]

limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)] = [(2) (-cos(0)2) – (4(0)2) sin (02)] / [(2(0)) (sec2(0)) tan (0) + (2) sec2 (0)]

limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)] = [(2) (1)2 – (4(0)(0)] / [(2(0)) (12) (0) + (2) (1)2]

limx→0 [2x cos(x)2] / [x sec2 (x) + (1) tan(x)] = [2 – 0] / [0 + 2] = 2/2 = 1

The limit of the function [sin (x2)] / [x tan (x)] as x approaches zero is 1.

## Example 11: Finding the Limit of a Complex Fraction

What is the limit of [3x + 10] / [2x2 – 6] as x approaches ∞?

Solution

First, identify the indeterminacy of the given equation by directly substituting the value of x = ∞. The equation is a ∞/∞ indeterminate form.

3(∞) + 10 = ∞

2(∞)2 - 6 = ∞

Apply the L’Hopital's Rule and differentiate the numerator and denominator separately.

limx→∞ [3x + 10] / [2x2 – 6] = limx→∞ [3 + 0] / [4x – 0]

limx→∞ [3x + 10] / [2x2 – 6] = limx→∞ 3 / 4x

limx→∞ [3x + 10] / [2x2 – 6] = 3 / ∞

limx→∞ [3x + 10] / [2x2 – 6] = 0

The limit of [3x + 10] / [2x2 – 6] as x approaches ∞ is 0.

## Example 12: Evaluating a Limit of Indeterminate Form 0/0 Multiplied by Negative Infinity

Find the limit limx→0 √x [ln (x)].

Solution

The given equation is a special indeterminate form besides 0/0 and ∞/∞. It is in an indeterminate form 0(-∞). Start by flipping √x.

limx→0+ √x [ln (x)]

limx→0+ [ln (x)] / [1/√x] = - ∞ / ∞

Apply the L’Hopital's Rule.

limx→0+ [ln (x)] / [1/√x] = limx→0+ [1/x] / [1/2x3/2]

limx→0+ [ln (x)] / [1/√x] = limx→0+ -2√x

limx→0+ [ln (x)] / [1/√x] = -2√0

limx→0+ [ln (x)] / [1/√x] = -2(0)

limx→0+ [ln (x)] / [1/√x] = 0

The limit of limx→0 √x [ln (x)] is 0.

## Example 13: Limit of an Inverse Trigonometric Equation

Calculate the limit limx→0 [arcsin (4x)] / [arctan (5x)].

Solution

Identify the determinacy of the given inverse trigonometric equation by substituting the value of x = 0 to the equation.

limx→0 [arcsin (4x)] / [arctan (5x)] = limx→0 [arcsin (0)] / [arctan (0)] = 0/0

Apply the L’Hopital's Rule and differentiate separately the numerator an denominator. Use the Chain Rule and get the derivatives of the top and bottom of the fraction equation.

limx→0 [arcsin (4x)] / [arctan (5x)] = limx→0 [(4) (1 /√(1-(4x)2))] / [(5) (1 /(1+(5x)2))]

limx→0 [arcsin (4x)] / [arctan (5x)] = limx→0 [4 / √(1 – 16x2)][(1 + 25x2)/5]

limx→0 [arcsin (4x)] / [arctan (5x)] = [4/5] [(1+25(0)2 / 1-16(0)2]

limx→0 [arcsin (4x)] / [arctan (5x)] = [4/5] [(1+0)/(1-0)]

limx→0 [arcsin (4x)] / [arctan (5x)] = 4/5

The limit of [arcsin (4x)] / [arctan (5x)] as x approaches zero is 4/5.

## Example 14: Limit of a Logarithmic Function

Evaluate the limit limx→0 (x) (ln(x)).

Solution

The given logarithmic equation is in the 0(-∞) indeterminate form. In order to apply the L’Hopital's Rule, flip the "x" portion of the given equation.

limx→0 (x) (ln(x)) = limx→0 (ln(x)) / (1/x) = (ln(0))/(1/0) = -∞/∞

Apply the L’Hopital's Rule and differentiate the numerator and denominator.

limx→0 (x) (ln(x)) = limx→0 (1/x) / (-1/x2)

limx→0 (x) (ln(x)) = limx→0 (1/x) / (x2/-1)

limx→0 (x) (ln(x)) = limx→0 (-x) = 0