L'Hopital's Rule
What is L'Hopital's Rule?
Referring back to limits, suppose we are trying to analyze the following limit:
[latex]\Large{\lim_{x \rightarrow 3}\frac{x^2-9}{x-3}}[/latex]
If you remember, if we try to solve this limit by direct substitution, we will end up with an indeterminate form: [latex]\frac{0}{0}[/latex]
One way to solve limits of this form is to use algebraic manipulation, as we have already explored. By factoring out the numerator first and canceling out common terms, we can evaluate the limit to be 6.
However, we will now introduce L’Hopital’s Rule, which will become very handy in evaluating the limits for indeterminate forms. This rule is outlined by the following:
L'Hopital's Rule:
For indeterminate forms of the type [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex] then:
[latex]\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}[/latex]
What does this tell us? Well, the rule tells us that the limit of [latex]\frac{f(x)}{g(x)}[/latex] is equal to the limit of the quotient of their derivatives, or [latex]\frac{f'(x)}{g'(x)}[/latex]. Let us now apply L’Hopital’s Rule to solve the limit given.
As we know, solving by direct substitution gives the indeterminate form:
[latex]\large{\lim_{x \to a}\frac{x^2-9}{x-3} = \frac{3^2-9}{3-3} = \frac{0}{0}}[/latex]
Applying the rule and our knowledge of evaluating limits as well as derivatives, we have:
[latex]\large{\lim_{x \to 3}\frac{x^2-9}{x-3} = \lim_{x \to 3}\frac{\frac{d}{dx}(x^2-9)}{\frac{d}{dx}(x-3)}}[/latex]
[latex]\large{ = \lim_{x \to 3}\frac{2x}{1} = \frac{2(3)}{1} = 6}[/latex]
As you can see, we have evaluated the limit to be 6. Which is the same answer we got when we used algebraic manipulation. This shows how efficiently the L’Hopital’s Rule can be used to find the limits of indeterminate forms.
Example
Evaluate the following limit:
[latex]\Large{\lim_{x \to 4}\frac{x^2-16}{x-4}}[/latex]
Notice that if we try solving by direct substitution, we will end up with [latex]\large{\frac{0}{0}}[/latex].
Thus, L’Hopital’s Rule can be used.
Applying L’Hopital’s Rule, we have:
[latex]\large{\lim_{x \to 4}[\frac{x^2-16}{x-4}] = \lim_{x \to 4}[\frac{\frac{d}{dx}(x^2-16)}{\frac{d}{dx}(x-4)}]}[/latex]
[latex]\large{= \lim_{x \to 4}[\frac{2x}{1}] = \frac{2(4)}{1} = 8}[/latex]
Example
Evaluate the following limit:
[latex]\Large{\lim_{x \to 1}\frac{\ln(x)}{x-1}}[/latex]
Notice that if we try solving by direct substitution, we will end up with [latex]\large{\frac{0}{0}}[/latex]. Thus, L’Hopital’s Rule can be used.
Applying L’Hopital’s Rule, we have:
[latex]\large{\lim_{x \to 1}[\frac{\ln(x)}{x-1}] = \lim_{x \to 1}[\frac{\frac{d}{dx}(\ln(x))}{\frac{d}{dx}(x-1)}]}[/latex]
We know the derivative of ln(x) is [latex]\frac{1}{x}[/latex].
Then:
[latex]\large{= \lim_{x \to 1}[\frac{\frac{1}{x}}{1}] = \frac{\frac{1}{1}}{1} = 1}[/latex]
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