L’ Hopital Rule in Calculus

The L'Hopital rule uses derivatives of each function to solve the limit, which helps us evaluate the limits that result in an indeterminate form.

L'Hopital rule states that when the limit is applied to a fraction of two functions resulting in an indeterminate form then it is equal to the limit of the fraction formed by the individual derivatives of functions.

Indeterminate Forms

The indeterminate forms are the forms with two functions whose limits cannot be determined by putting the limits in the function. The indeterminate form is the form that is undefined mathematically.

The forms whose value cannot be evaluated by directly applying the limits are called indeterminate forms. Indeterminate form includes 0/0, ±∞/±∞, 0×∞, ∞-∞, 0⁰, 1^∞ etc. The 0/0, and ±∞/±∞ are the most common indeterminate forms in which the L'Hopital rule is applied.

L'Hopital Rule Formula

For two continuous and differentiable functions f(x) and g(x), if the limits x tends to result in an indeterminate form, then the L'Hopital rule is applied, and it states,

If the limit \lim_{{x \to c}} \frac{f(x)}{g(x)}​ results in an indeterminate form of \frac{0}{0}​ or \frac{\infty}{\infty}​, and

if the derivatives f′(x) and g′(x) exist and

are continuous near c, and g′(x)≠0 for x near c, then:

\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}

Where,

• a is any real number or infinity.
• f'(x) is derivative of f(x)
• g'(x) is derivative of g(x) and g(x) and g(a) ≠ 0

Conditions for L'Hopital Rule

The L Hopital rule is used when the limits of two differentiable functions after applying the limit gives an indeterminate form. Commonly, for the indeterminate forms 0/0, ±∞/±∞ we apply the L'Hopital rule directly to evaluate the limit.

Some necessary conditions for applying the L'Hopital rule

• f(x) and g(x) must be differentiable.
• The limit of the quotient of the derivatives of a given function should exist i.e.,

\lim_{{x \to a}} \frac{f'(x)}{g'(x)}= Some Finite Number

L'Hopital Rule Proof

The L'Hopital rule is applied when limits result in indeterminate form 0/0, ±∞/±∞. We can prove the L'Hopital rule by using Cauchy's Mean Value Theorem.

Let f(x) and g(x) be two continuous functions on the interval [a, b] and differentiable on the interval (a, b) and we know for any function f(x), its derivative at x = c is given as

f'(c) = \frac{f(x) - f(c)}{(x - c)}

Assume that g(x) = 0 on (a, b), then there exists c in (a, b) such that

\lim_{{x\to b}} \frac{f'(x)} {g'(x)} = \lim_{{ x \to b}} \frac{(f(x)- f(b)) / (x - b)} {(g(x)- g(b)) / (x - b)}

Let the functions f and g be differentiable at x = c satisfying where c belongs to the interval in which functions are defined. Let f(c) = g(c) = 0.

By Cauchy's Mean Value Theorem,

\lim_{{x\to c}} \frac{f(x)}{g(x)}= lim _x→c[f(x) -f(c)]/[g(x) - g(c)] \lim_{{x\to c}}\frac {f(x) -f(c)}{g(x) - g(c)}

⇒ \lim_{{x\to c}} \frac{f(x)}{g(x)}=\lim_{{x\to c}}\frac {f(x) - 0} {g(x) - 0}   [as f(c) = g(c) = 0]

⇒ \lim_{{x\to c}} \frac{f(x)}{g(x)}= \lim_{{x\to c}}\frac{f(x) - f(c)}{g(x) - g(c)}

⇒ \lim_{{x\to c}} \frac{f(x)}{g(x)}= \lim_{{x\to c}}\frac{\frac{(f(x)- f(c))}{(x - c)}}{\frac{(g(x)- g(c))}{(x - c)}}

⇒ \lim_{{x\to c}} \frac{f(x)}{g(x)} = \lim_{{x\to c}} \frac{f'(x)}{g'(x)}

Which is the required result.

How to Apply L'Hopital Rule?

L'Hopital rule is applied when we get an indeterminate form after applying the limits. For applying the L'Hopital Rule these are the following steps:

• First check after applying the limit we get indeterminate form or not. If we do not get indeterminate form then, we cannot apply L'Hopital rule.
• When we get indeterminate form, simplify the fraction before applying the rule.
• Find the derivatives of the functions separately.
• Determine the value of ratio of derivatives to get the result.

Example: Find lim_x→1 [(x¹⁰ - 1) / (x² - 1)]

Solution:

First apply the limit

lim_x→1 [(x¹⁰ - 1) / (x² - 1)] = [1¹⁰ -1] / [1² -1] = 0 / 0

After applying the limit, we get 0 / 0 form so, we apply L'Hopital rule

f(x) = x¹⁰ - 1

⇒ f'(x) = 10x⁹

g(x) = x² - 1

⇒ g'(x) = 2x

lim_x→1 [(x¹⁰ - 1) / (x² - 1)] = (10x⁹) / 2x

Now, put limit x = 1, we get

⇒ lim_x→1 [(x¹⁰ - 1) / (x² - 1)] = [(10×1⁹ )-1] / [2×1] = 10 / 2

⇒ lim_x→1 [(x¹⁰ - 1) / (x² - 1)] = 5

Applying L'Hopital Rule Multiple Times

When the L'Hopital rule is applied once and also after applying the limit we get an indeterminate form then after simplifying the fraction, we can again apply L'Hopital Rule. We can apply the rule multiple times until we get an indeterminate form. Before each application of the L'Hopital rule, we should simplify the fractions.

Example: Find lim_x→0 [(sinx - x) / x²]

Solution:

First apply the limit

lim_x→0 [(sinx - x) / x²] = (sin0 - 0) / 0² = 0/0

After applying the limit, we get 0 / 0 form so, we apply L'Hopital rule

f(x) = sin x - x

⇒ f'(x) = cos x -1

g(x) = x²

⇒ g'(x) = 2x0

lim_x→0 [(sinx - x) / x²] = (cos x -1) / 2x

Now, put limit x = 0, we get

lim_x→0 [(sinx - x) / x²] = (cos 0 -1) / 2×0 = 0 / 0

Again we get 0 / 0 form so, we again apply L'Hopital rule

f(x) = cos x - 1

⇒ f'(x) = -sinx

g(x) = 2x

⇒ g'(x) = 2

lim_x→0 [(sinx - x) / x²] = (-sin x) / 2

Now, put limit x = 0, we get

lim_x→0 [(sinx - x) / x²] = (- sin 0) / 2 = 0

⇒ lim_x→0 [(sinx - x) / x²] = 0

How to apply L'Hopital Rule for Exponents

When the variable is in power, we equate the limit with a variable and take the logarithm on both sides and then apply the L'Hopital rule.

Example: Evaluate: lim_x→∞ x^1/x

Solution:

The given limit gives ∞⁰

In this question we have variable in the exponent, so we take ln

y = x^1/x

Taking ln both sides in the above equation.

ln y = ln x^1/x

⇒ ln y = (1 / x) ln x

Applying limit

lim_x→∞ ln y = lim_x→∞ (1 / x) × ln x

⇒ lim_x→∞ ln y = lim_x→∞ (ln x) / x

Applying L'Hopital rule

lim_x→∞ ln y = lim_x→∞ (1 / x) / 1

⇒ lim_x→∞ ln y = lim_x→∞ (1 / x)

putting limit

lim_x→∞ ln y = 1 / ∞ = 0

⇒ ln [lim_x→∞ y] = 0

⇒ lim_x→∞ y = e⁰

⇒ lim_x→∞ y = 1

⇒ lim_x→∞ x^1/x= 1

Key Points about L’ Hopital Rule

Some of the key points which we need to remember related to L'Hopital's rule are:

• If the limit does not result in an indeterminate form after applying the limit, then we cannot apply the L'Hopital rule (it provides an incorrect result).
• Each time before applying the rule, we should simplify the rational expression.
• If the limit is given in product form like f(x). g(x) then, for applying the rule we can rewrite it as f(x) / 1 / g(x) or g(x) / 1 / f(x).
• While applying the L'Hopital rule multiple times always check if the resulting rational expression gives an indeterminate form or not.

Also Check

• Calculus
• Introduction to Limits
• Sandwich Theorem
• Indeterminate Forms
• Calculus Formulas

Solved Examples with Solutions

Example 1. Find lim_x→-2 [(x + 2) / (x² + 3x + 2)]

Solution:

First apply the limit

lim_x→-2 [(x + 2) / (x² + 3x + 2)] = lim_x→-2 [(- 2 + 2) / ((-2)² + 3×(-2) + 2)] = 0 / 0

After applying the limit, we get 0 / 0 form so, we apply L'Hopital rule

f(x) = x + 2

⇒ f'(x) = 1

g(x) = x² + 3x + 2

⇒ g'(x) = 2x + 3

lim_x→-2 [(x + 2) / (x² + 3x + 2)] = 1 / (2x + 3)

Now, put limit x = -2, we get

lim_x→-2 [(x + 2) / (x² + 3x + 2)] = 1 / [2×(-2) + 3] = 1 / (-4 + 3)

⇒ lim_x→-2 [(x + 2) / (x² + 3x + 2)] = -1

Example 2. Find lim_x→0 [sin 3x / x]

Solution:

First apply the limit

lim_x→0 [sin 3x / x] = sin 3(0) / 0 = 0 / 0

After applying the limit, we get 0 / 0 form so, we apply L'Hopital rule

f(x) = sin3x

⇒ f'(x) = 3cos3x

g(x) = x

⇒ g'(x) = 1

lim_x→0 [sin 3x / x] = 3cos 3x / 1

Now, put limit x = 0, we get

lim_x→0 [sin 3x / x] = [3cos 3(0)] / 1 = 3 / 1

⇒ lim_x→0 [sin 3x / x] = 3

Example 3. Find lim_x→∞ [x² / e^x]

Solution:

First apply the limit

lim_x→∞ [x² / e^x] = ∞² / e ^∞ = ∞ / ∞

After applying the limit, we get ∞ / ∞ form so, we apply L'Hopital rule

f(x) = x²

⇒ f'(x) = 2x

g(x) = e^x

⇒ g'(x) = e^x

⇒ lim_x→∞ [x² / e^x] = 2x / e^x

Now, put limit x = ∞, we get

lim_x→∞ [x² / e^x] = (2 × ∞)/ e^∞ = ∞ / ∞

Again we get ∞ / ∞ form so, we again apply L'Hopital rule

f(x) = 2x

⇒ f'(x) = x

g(x) = e^x

⇒ g'(x) = e^x

lim_x→∞ [x² / e^x] = 2 / e^x

Now, put limit x = ∞, we get

lim_x→∞ [x² / e^x] = 2 / e^x = 2 / e^∞ = 2 / ∞

⇒ lim_x→∞ [x² / e^x] = 0

Example 4. Evaluate: lim_x→∞ x^x

Solution:

The given limit gives ∞^∞

In this question we have variable in the exponent, so we take ln

y = x^x

Taking ln

ln y = ln xˣ

⇒ ln y = x ln x

Applying Limit

limₓ→∞ ln y = limₓ→∞ x ln x

⇒ limₓ→∞ ln y = limₓ→∞ [ ln x / (1 / x) ]

Applying L'Hopital rule

limₓ→∞ ln y = limₓ→∞ [ (1 / x) / ( −1 / x² ) ]

⇒ limₓ→∞ ln y = limₓ→∞ [ −(1 / x) / (1 / x)² ]

⇒ limₓ→∞ ln y = limₓ→∞ [ −1 / (1 / x) ]

⇒ limₓ→∞ ln y = limₓ→∞ ( −x )

Since for large x,
x > 0 and ln x > 0,
therefore x ln x is positive and increases without bound.

Hence,

limₓ→∞ ln y = +∞

⇒ ln ( limₓ→∞ y ) = +∞

⇒ limₓ→∞ y = e⁺∞

⇒ limₓ→∞ y = ∞

⇒ limₓ→∞ xˣ = ∞

Practice Problems on L Hopital Rule

Find the following limits using L' Hopital Rule

Problem: 1\lim_{{x \to 2}} \frac{{x^2 - 4}}{{x - 2}}

Problem: 2\lim_{{x \to 0}} \frac{{e^x - 1}}{{x}}

Problem: 3\lim_{{x \to \infty}} \frac{{x^2 + 3x}}{{2x^2 - 5}}

Problem: 4\lim_{{x \to \infty}} \frac{{\ln(x)}}{{x}}

Problem: 5\lim_{{x \to 0}} \frac{{\sin(x)}}{{x}}

Problem: 6\lim_{{x \to 0}} \frac{{1 - \cos(x)}}{{x^2}}

Problem: 7\lim_{{x \to \infty}} \left(1 + \frac{2}{x}\right)^x

Problem: 8\lim_{{x \to \infty}} \left(\frac{{x + 1}}{{x}}\right)^{3x}