In our study of limits, we've often encountered limits that we can't evaluate just by plugging in the number. For example, $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$. Plugging in $x=2$ gives the form $\frac{0}{0}$, which is meaningless.
So far, our strategy has been purely algebraic: we factor, rationalize, or use other tricks to rewrite the expression. For example:
But what about a limit like $\lim_{x \to 0} \frac{\sin x}{x}$? We know this limit is 1, but we had to use a complex geometric argument (the Squeeze Theorem) to prove it. Algebra alone doesn't work. This limit also has the form $\frac{0}{0}$.
Today, we introduce a powerful new tool that uses derivatives to solve these kinds of limits systematically. This tool is called L'Hôpital's Rule.
An indeterminate form is a limit expression that does not provide enough information to determine its value. The form $\frac{0}{0}$ is called indeterminate because the limit could be anything—it could be 4, it could be 1, or it could be $\infty$. The final value depends on the relationship between the numerator and the denominator as they both approach zero.
There are several types of indeterminate forms. L'Hôpital's Rule applies directly to two of them:
The other indeterminate forms can all be algebraically manipulated to become one of these two quotient forms. They are:
Be careful! Not all forms involving 0 or $\infty$ are indeterminate. For example:
You can only use L'Hôpital's Rule on the indeterminate forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
L'Hôpital's Rule (L'H) provides the tool we were missing. It says that if a limit has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can take the derivative of the numerator and the derivative of the denominator separately and then try the limit again.
Suppose $f$ and $g$ are differentiable and $g'(x) \neq 0$ near $a$ (except possibly at $a$). Suppose that $$\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0$$ OR $$\lim_{x \to a} f(x) = \pm\infty \quad \text{and} \quad \lim_{x \to a} g(x) = \pm\infty$$
Then: $$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$
This rule also applies to one-sided limits and limits as $x \to \infty$ or $x \to -\infty$.
Why does this work? Think of the $\frac{0}{0}$ case. Both $f(x)$ and $g(x)$ are "racing" to zero. The value of the limit is determined by who wins the race. Derivatives measure rates of change—or in this case, speed.
If $f(x)$ is heading to zero *faster* than $g(x)$, the limit will be 0. If $g(x)$ is faster, the limit will be $\infty$. If they are moving at comparable speeds, the limit will be some finite number, which is the ratio of their speeds (their derivatives).
Near $x=a$, we can use linear approximation: $f(x) \approx f(a) + f'(a)(x-a) = 0 + f'(a)(x-a)$ $g(x) \approx g(a) + g'(a)(x-a) = 0 + g'(a)(x-a)$
So, the ratio looks like:
This isn't a formal proof, but it gives the correct intuition: the limit of the ratio of the functions is equal to the limit of the ratio of their rates.
L'Hôpital's Rule states that $\lim \frac{f}{g} = \lim \frac{f'}{g'}$.
It does NOT state that $\lim \frac{f}{g} = \lim \left(\frac{f}{g}\right)'$. The derivative of a quotient is $\frac{g f' - f g'}{g^2}$, which is completely different. You are differentiating the top and bottom separately.
Find $\lim_{x \to 0} \frac{\sin x}{x}$.
Step 1. Check the form. Plugging in $x=0$, we get $\frac{\sin 0}{0}$, which is $\frac{0}{0}$. This is an indeterminate form, so L'Hôpital's Rule applies.
Step 2. Apply L'Hôpital's Rule. We differentiate the numerator and denominator separately.
$$ \lim_{x \to 0} \frac{\sin x}{x} \overset{L'H}{=} \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} $$ $$ = \lim_{x \to 0} \frac{\cos x}{1} $$Step 3. Evaluate the new limit. Now we can plug in $x=0$.
$$ \lim_{x \to 0} \frac{\cos x}{1} = \frac{\cos 0}{1} = \frac{1}{1} = 1 $$This confirms our previous result from the Squeeze Theorem, but with much less effort!
Find $\lim_{x \to \infty} \frac{e^x}{x^2}$. (Which "wins": the exponential or the polynomial?)
Step 1. Check the form. As $x \to \infty$, $e^x \to \infty$ and $x^2 \to \infty$. The form is $\frac{\infty}{\infty}$. L'Hôpital's Rule applies.
Step 2. Apply L'Hôpital's Rule.
$$ \lim_{x \to \infty} \frac{e^x}{x^2} \overset{L'H}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(x^2)} = \lim_{x \to \infty} \frac{e^x}{2x} $$Step 3. Evaluate the new limit. As $x \to \infty$, we check the form *again*. The new limit is $\frac{e^\infty}{2\infty}$, which is still $\frac{\infty}{\infty}$!
Step 4. Apply L'Hôpital's Rule again. Since the new limit is also indeterminate, we can apply the rule a second time.
$$ \lim_{x \to \infty} \frac{e^x}{2x} \overset{L'H}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}(e^x)}{\frac{d}{dx}(2x)} = \lim_{x \to \infty} \frac{e^x}{2} $$Step 5. Evaluate the final limit. Now, as $x \to \infty$, the numerator $e^x \to \infty$ while the denominator is a constant 2. $$ \lim_{x \to \infty} \frac{e^x}{2} = \frac{\infty}{2} = \infty $$
Conclusion: $\lim_{x \to \infty} \frac{e^x}{x^2} = \infty$. This shows that the exponential function $e^x$ grows "infinitely faster" than the polynomial $x^2$.
Find the limit: $\lim_{x \to 1} \frac{1 - x + \ln x}{x^3 - 3x + 2}$
What about the other forms? The trick is to use algebra to turn them into $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
The trick is to invert and divide. If you have $\lim f(x)g(x)$ where $f \to 0$ and $g \to \infty$, you can rewrite it as:
OR
One of these is usually easier to differentiate than the other. Choose the simpler one!
Find $\lim_{x \to 0^+} x \ln x$.
Step 1. Check the form. As $x \to 0^+$, we have $x \to 0$ and $\ln x \to -\infty$. The form is $0 \cdot (-\infty)$, an indeterminate product.
Step 2. Rewrite as a quotient. We have two choices. Choice A: $\frac{x}{1/\ln x}$. (This is $\frac{0}{0}$. But the derivative of $1/\ln x$ is messy.) Choice B: $\frac{\ln x}{1/x}$. (This is $\frac{-\infty}{\infty}$. The derivatives are simple. Let's use this one!)
$$ \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} $$Step 3. Apply L'Hôpital's Rule. The form is $\frac{-\infty}{\infty}$, so L'H applies.
$$ \lim_{x \to 0^+} \frac{\ln x}{1/x} \overset{L'H}{=} \lim_{x \to 0^+} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x^{-1})} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} $$Step 4. Simplify and Evaluate. Don't use L'H again! Just simplify the algebra.
$$ \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \left( \frac{1}{x} \cdot \frac{-x^2}{1} \right) = \lim_{x \to 0^+} (-x) $$This limit is no longer indeterminate. $$ \lim_{x \to 0^+} (-x) = 0 $$
The trick is to find a common denominator or use other algebraic methods (like rationalizing or factoring) to combine the terms into a single quotient.
Find $\lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{e^x - 1} \right)$.
Step 1. Check the form. As $x \to 0^+$, $\frac{1}{x} \to \infty$. As $x \to 0^+$, $e^x - 1 \to e^0 - 1 = 0$, so $\frac{1}{e^x - 1} \to \infty$. The form is $\infty - \infty$.
Step 2. Rewrite as a quotient. We find a common denominator.
$$ \lim_{x \to 0^+} \left( \frac{e^x - 1}{x(e^x - 1)} - \frac{x}{x(e^x - 1)} \right) = \lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)} $$Step 3. Check the new form. Plug in $x=0$. Numerator: $e^0 - 1 - 0 = 1 - 1 = 0$. Denominator: $0(e^0 - 1) = 0$. The new form is $\frac{0}{0}$. L'Hôpital's Rule applies.
Step 4. Apply L'Hôpital's Rule. We need the product rule for the denominator.
$$ \lim_{x \to 0^+} \frac{e^x - 1 - x}{x e^x - x} \overset{L'H}{=} \lim_{x \to 0^+} \frac{e^x - 1}{(x e^x + e^x \cdot 1) - 1} = \lim_{x \to 0^+} \frac{e^x - 1}{x e^x + e^x - 1} $$Step 5. Check the form *again*. Plug in $x=0$. Numerator: $e^0 - 1 = 0$. Denominator: $0 \cdot e^0 + e^0 - 1 = 0 + 1 - 1 = 0$. The form is *still* $\frac{0}{0}$. We apply L'Hôpital's Rule again.
Step 6. Apply L'Hôpital's Rule again.
$$ \lim_{x \to 0^+} \frac{e^x - 1}{x e^x + e^x - 1} \overset{L'H}{=} \lim_{x \to 0^+} \frac{e^x}{(x e^x + e^x \cdot 1) + e^x - 0} = \lim_{x \to 0^+} \frac{e^x}{x e^x + 2e^x} $$Step 7. Evaluate the final limit. Plug in $x=0$.
$$ \frac{e^0}{0 \cdot e^0 + 2e^0} = \frac{1}{0 + 2(1)} = \frac{1}{2} $$Our last category involves powers. These forms are tricky because the base and the exponent are "fighting" each other. (e.g., in $0^0$, the base $0$ wants the answer to be 0, but the exponent $0$ wants the answer to be 1).
This is a great question. Let's think about the "tug-of-war" for a limit of the form $\lim [f(x)]^{g(x)}$ where $f(x) \to 0$ and $g(x) \to \infty$.
What happens when you take a tiny number (less than 1) and raise it to a huge power? $(0.1)^{10}$ is tiny. $(0.01)^{100}$ is unimaginably tiny.
Unlike a form like $1^\infty$ (where the base and exponent "fight" each other), here both forces are working together. The base is pulling the result to $0$, and the infinitely large exponent is just helping it get to $0$ even faster.
There is no ambiguity or "fight." The result is always $0$. Therefore, $0^\infty$ is a determinate form, and its value is $0$. We don't need L'Hôpital's Rule for it.
The strategy for the truly indeterminate powers is to use the power of logarithms to turn the problem into a product.
Find $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$. (This is the definition of $e$!)
Step 1. Check the form. As $x \to \infty$, $1/x \to 0$. The base $(1 + 1/x) \to 1$. The exponent is $\infty$. The form is $1^\infty$.
Step 2. Use the Log-Trick. Let $y = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$. Take the log of both sides:
Step 3. Check the new form. This is now $\infty \cdot \ln(1+0) \to \infty \cdot 0$. An indeterminate product.
Step 4. Rewrite as a quotient. We will move the $x$ to the denominator as $1/x$.
Step 5. Apply L'Hôpital's Rule. Let's be careful with the Chain Rule.
Step 6. Simplify and Evaluate. The $-x^{-2}$ terms cancel beautifully!
Step 7. Solve for y (The Final, Critical Step). We have successfully found that $\ln y = 1$. This is the limit of the logarithm of our answer. We need to find $y$ itself. To undo the $\ln()$, we must use its inverse function, which is $e^x$. We will exponentiate both sides of the equation:
By the fundamental inverse property of logs and exponentials, $e^{\ln y}$ simplifies to just $y$.
So, we have our final answer: $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$.
Find the limit: $\lim_{x \to 0^+} x^x$
L'Hôpital's Rule is a powerful technique for evaluating limits that are indeterminate. The key is to first identify which form you have, then use the correct strategy.
| Indeterminate Form | Example | Strategy |
|---|---|---|
| $\frac{0}{0}$ or $\frac{\infty}{\infty}$ | $\lim_{x \to 0} \frac{\sin x}{x}$ | Apply L'Hôpital's Rule directly: $\lim \frac{f'}{g'}$. |
| $0 \cdot \infty$ | $\lim_{x \to 0^+} x \ln x$ | Rewrite as a quotient: $\lim \frac{f}{1/g}$ or $\lim \frac{g}{1/f}$. Then use L'H. |
| $\infty - \infty$ | $\lim_{x \to 0} (\frac{1}{x} - \frac{1}{\sin x})$ | Find a common denominator to create a single quotient. Then use L'H. |
| $1^\infty, 0^0, \infty^0$ | $\lim_{x \to 0^+} x^x$ | Use the "Log-Trick." Let $y = \lim[\dots]$, find $\lim(\ln y)$, then solve for $y = e^{\text{limit}}$. |
After this lecture, you should be able to: