Applying L'Hôpital's Rule

The limit of a function may not be obvious such as a limit that arrives at 0(∞) \( {0\over 0}\), \( {∞\over ∞}\), \( 0^{0}\),\( ∞^{∞}\) or \( 1^{∞}\). L'Hôpital's Rule is a tool that is used to find the limit of the functions that are in indeterminate form. L'Hôpital's Rule is used for rational functions as we are required to differentiate the numerator and denominator until we find a limit which is either a finite value or infinity. You need to set up the function as a quotient and then afterwards differentiate the numerator and denominator until you arrive at a determinate limit. However, the limit may be a product or in exponent form so we will need to use logarithms first before we can use L'Hôpital's Rule. We will first look at the proof of L'Hôpital's Rule to understand why it can only be used for indeterminate limits and then explore its application.

Limit of Functions in Indeterminate form

Proof of L'Hôpital's Rule

It is important to review the proof of L'Hôpital's Rule in order to understand why it can only be used for limits that are in indeterminate form
f(b) = g(b) = 0
Lim _x→b \( {f(x)\over g(x)}\)= Lim _x→b \( {f(x)-f(b)\over g(x)-g(b)}\) = Lim _x→b = \( {{f(x)-f(b)\over x-b}\over {g(x)-g(b)\over x-b}}\)
Using the quotient rule of limits
Lim _x→b \( {f(x)\over g(x)}\) = \( {\text{Lim}_{x→b}{f(x)-f(b)\over x-b}\over \text{Lim}_{x→b}{g(x)-g(b)\over x-b}}\)
Applying the first principles of differentiation leads to:
Lim _x→b \( {f(x)\over g(x)}\) = \({f'(b)\over g'(b)}\)
Lim _x→b f(x) = Lim _x→b g(x) = ∞
L = Lim _x→b \({f(x)\over g(x)}\) = Lim \( { {1\over g(x)}\over {1 \over f(x)}}\)
Let us show that L can also be expressed as the following:
L = Lim _x→b \({f(x)\over g(x)}\) = Lim \( { ({1\over g(x)})'\over ({1 \over f(x)})''}\) = Lim _x→b \({-(g(x))^{-2}g'(x)\over -(f(x))^{-2}f'(x)}\)
Lim _x→b \({(f(x))^{2}g'(x)\over (g(x))^{2}f'(x)}\)
Next we will use the product and power properties of limits.
L = L² Lim _x→b \({g'(x)\over f'(x)}\)
Hence, we have shown that :

L = Lim _x→b \({f(x)\over g(x)}\) = Lim _x→b \({f'(x)\over g'(x)}\)
L'Hôpital's Rule can only be applied to quotients in the form of zero over zero or infinity over infinity. Next we will look at different examples of indeterminate form and show how we can use L'Hôpital's Rule correctly to find the limit.

Finding the limit of \( {0\over 0}\)

Example- Finding lim _x→0 \( {sin(x)\over x}\)
Here we can apply L'Hôpital's Rule by differentiating both the numerator and the denominator.
lim _{x→ 0} \( {sin(x)\over x}\) = lim _x→0 \( {cos(x)\over x}\) = 1

Finding the limit of \( {∞\over ∞}\)

Example- Finding lim _{x→ ∞} \( {2x+4\over x+3}\)
Here we can apply L'Hôpital's Rule rule by differentiating both the numerator and the denominator.
lim _x→∞\( {2x+4\over x+3}\) = lim _x→∞ \( {2\over 1}\) = 2

Finding the limit of 0⁰

Example- Finding lim _x→0+ x^tan(x)
In order to find the limit we need to express it as a quotient. We will define this limit as the following:
lim _x→0+ x^tan(x) = L
Let us apply natural logarithms.
lim _x→0+ln x^tan(x) = ln(L) = A
Using properties of logarithms.
lim _x→0+ tan(x)ln(x) = A
We wish to find A, but firstly let us write it as a quotient.
lim _x→0+ \( {ln(x)\over (tan(x))^{-1}}\) = lim_{x--> 0+} \( {(1/x)\over -1(tan(x))^{-2}(sec(x))^{2}}\)
lim _x→0+ \( {ln(x)\over (tan(x))^{-1}}\)= -lim _{x--> 0+} \( {sin(x)\over x}\) lim _{x--> 0+} sin(x)
lim _x→0+ \( {ln(x)\over (tan(x))^{-1}}\)= 1(0) = 0
lim _x→0+ tan(x)ln(x)= 0 = A
Since we know A, we can find L.
ln A = L = 0
L = e⁰= 1
Hence,
lim _{x→ 0+} x^tan(x)= 1

Finding the limit of 1^∞

Example- Finding lim _x→∞ \(( 1 + {1\over x})^{x}\)
In order to find the limit we need to express it as a quotient. We will define this limit as the following:
lim _x→∞ \( (1 + {1\over x})^{x}\) = L
Let us apply natural logarithms.
lim _x→∞ln (\( 1 + {1\over x})^{x}\) = ln(L) = A
Using properties of logarithms:
lim _x→∞ xln(\( 1 + {1\over x})\) = A
We wish to find A, but firstly let us write it as a quotient.
lim _x→∞ \( {ln( 1 + (x)^{-1})\over (x)^{-1}}\)
Now let us differentiate both the numerator and the denominator.
lim _x→∞ \( { -1(x)^{-2} ( 1 + (x)^{-1})^{-1}\over (-1)(x)^{-2}}\)
Next we will need to simplify:
lim _x→∞ \( ( 1 + (x)^{-1})^{-1}\)
Then finally apply limits to show that:
A = lim _x→∞ \( ( 1 + (x)^{-1})^{-1}\) = 1
Since ln(L) = A, we can conclude that L = e.

Finding the limit of ∞^∞

Example- Finding lim _x→∞ \((x)^{x}\)
In order to find the limit we need to express it as a quotient. We will define this limit as the following:
lim _x→∞ \((x)^{x}\) = L
Let us apply natural logarithms.
lim _x→∞ln \((x)^{x}\) = ln(L) = A
Using properties of logarithms.
lim _x→∞ xln(x) = A
Here we do not apply L'Hôpital's Rule since we can show that it is not in indeterminate form. Using the multiplicative property of limits.
lim _x→∞ xln(x) = lim _x→∞ x lim _x→∞ ln(x) = (∞)(∞)
Since ln (L) = A. We can conclude that L = ∞

Finding the limit of 0^∞

Example- Finding lim _x→∞ (\({1\over x})^{x}\)
In order to find the limit we need to express it as a quotient. We will define this limit as the following:
lim _x→∞ (\({1\over x})^{x}\)= L
Let us apply natural logarithms.
lim _x→∞ln lim _x→∞ (\({1\over x})^{x}\) = ln(L) = A
Using properties of logarithms.
lim _x→∞ -xln(x) = A
Here, we do not use L'Hôpital's Rule since it is not in indeterminate form. Using the multiplicative property of limits.
lim _x→∞ -xln(x) = -lim _x→∞ x lim _x→∞ ln(x) = -(∞)(∞)
Since ln (L) = A. We can conclude that L = 0.

Repeated Application of L'Hopital's Rule

Applying L'Hôpital's Rule once may not be enough as we still have a limit in indetermite form. As a result, we will need to continue to apply L'Hôpital's Rule until our limit is finite
Example- Finding lim _{x→ ∞} \( {2x^2+4x\over x^2+3}\)
Differentiating once and applying limits still leads to an indeterminate value
lim _{x→ ∞} \( {4x+4\over 2x}\) = \( {0\over 0}\)
Hence we will need to differentiate again and can show that this time the limit is finite.
lim _{x→ ∞} \( {4\over 2}\) = 2

Key Takeaways

L'Hôpital's Rule is a method for finding a limit in indeterminate form. The function should be expressed as a quotient so that we can differentiate the numerator and denominator until we find the limit. If the limit is written as a product or is in index form then logarithms can be applied. L'Hôpital's Rule can only be used with quotients so there are times when we may need to apply logarithms to express it as a rational function. Once these two requirements are met we can apply L'hopital repeatedly until we have a determinate value.