Basically, the rule says that for certain types of limits called indeterminate forms (more on that below), the limit of a ratio of functions is equal to the limit of the ratio of their derivatives. Here it is:

You'll find a couple of examples of those kinds of manipulations in the examples that follow.

$$\infty \cdot \infty \phantom{0000} \infty \cdot 0 \phantom{0000} 0^{\infty}$$

$$0^0 \phantom{0000} \infty^0$$

A proof of L'Hôpital's rule is included below, but for now it's best to learn to use it by working examples. L'Hôpital's is a terrific time-saver.

Caution: one common mistake is to apply the quotient rule. We're not taking the derivative of a quotient here. We're finding the derivatives of the numerator and denominator separately – no quotient rule.

We can now do a little algebra and plug $x = 0$ in to find that this function "blows up." It tends toward infinity as $x$ approaches zero. The cosine term will oscillate between zero and one, but the square root term will be zero when $x = 0$.

$$\lim_{x\to 1} \, \frac{x^2 - 1}{ln(x)} = \lim_{x\to 1} \frac{2x}{\frac{1}{x}}$$

Now we can rearrange (division by a fraction is the same as multiplication by the reciprocal) and substitute $x = 1$ to get:

$$\lim_{x\to 1} \, 2x^2 = 2$$

So this function has a limit of $2$ at $x = 1$.

Taking the derivative once again gives us an   $\frac{\infty}{\infty}$   limit:

$$\lim_{x\to\infty} \, \frac{6x^2 - 10x}{3x^2} = \lim_{x\to\infty} \, \frac{12x - 10}{6x}$$

But that's still OK; the rule still applies, so we'll take another round of derivatives to get:

$$\lim_{x\to\infty} \, \frac{12x - 10}{6x} = \lim_{x\to\infty} \, \frac{12}{6} = 2$$

Finally, we have an answer. As $x \rightarrow \infty$, the function approaches a fixed limit of 2.

Now we can take derivatives as usual, to get:

$$\lim_{x\to 0^+} \, \frac{\frac{1}{x}}{-x^{-2}} = \lim_{x\to 0^+} \, (-x) = 0$$

We need to take a minute to consider that this is a one-sided limit. It's actually a natural way to write a limit like this; recall that the domain of a function like this, containing a log function, is $x \ge 0$.

$$ \begin{align} &= \lim_{x\to 0} \, \frac{x - \text{sin}(x)}{x \text{sin}(x)} \\[5pt] &= \lim_{x\to 0} \, \frac{1 - \text{cos}(x)}{x \text{cos}(x) + \text{sin}(x)} \end{align}$$

So we just apply the rule again to get a limit we can finally evaluate,

$$= \lim_{x\to 0} \, \frac{\text{sin}(x)}{-x \text{sin}(x) + 2 \text{cos}(x)} = 0$$

... so you may have to try a few things (some of which may fail – no big deal – and be persistent about applying L'Hôpital's rule to solve some of these limits.

$$ \begin{align} \lim_{x\to \infty} \, ln(y) &= \lim_{x\to \infty} \, \frac{ln(x)}{x} \\[5pt] &= \lim_{x\to \infty} \, \frac{1/x}{1} = 0 \end{align}$$

Notice that the second term in that last equation series is an $\frac{\infty}{\infty}$ limit. Now we apply the inverse function, $e^y$, to our result to get the limit:

$$ \begin{align} \lim_{x\to \infty} \, y &= \lim_{x\to \infty} \, e^{ln(y)} \\[5pt] &= e^{\lim_{x\to \infty} \, ln(y)} \\[5pt] &= e^0 = 1 \end{align}$$

9. $$\lim_{x\to 1} \frac{x^3 - x^2}{x^3 - 1}$$

   Solution

   $$ \begin{align} \lim_{x\to 1} \, \frac{x^3 - x^2}{x^3 - 1} \; &\longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 1} \, \frac{3x^2 - 2x}{3x^2} \\[5pt] &= \frac{3 - 2}{3} = \frac{1}{3} \end{align}$$




10. $$\lim_{x\to 0} \frac{\text{tan}(x)}{\text{tan}(3x)}$$

    Solution

    $$ \begin{align} \lim_{x\to 0} \, \frac{\text{tan}(x)}{\text{tan}(3x)} \; &\longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 0} \, \frac{sec^2(x)}{3 sec^2(3x)} = \lim_{x\to 0} \, \frac{\frac{1}{\text{cos}^2(x)}}{\frac{3}{\text{cos}^2(3x)}} \\[5pt] &= \lim_{x\to 0} \, \frac{\text{cos}^2(3x)}{3 \text{cos}^2(x)} = \frac{\text{cos}^2(0)}{3 \text{cos}^2(0)} = \frac{1}{3} \end{align}$$




11. $$\lim_{x\to\infty} x^3 e^{-x}$$

    Solution

    $$ \begin{align} \lim_{x\to\infty} \, x^3 e^{-x} &= \lim_{x\to\infty} \frac{x^3}{e^x} \; \longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{3x^2}{e^x} = \lim_{x\to\infty} \frac{6x}{e^x} \\[5pt] &= \lim_{x\to\infty} \frac{6}{e^x} = 0 \end{align}$$




12. $$\lim_{x\to 0} x^2 \, ln(x)$$

    Solution

    $$\lim_{x\to 0} \, x^2 \, ln(x)$$

    We can manipulate this limit into an $\frac{\infty}{\infty}$ limit like this:

    $$ \begin{align} \lim_{x\to 0} \, \frac{ln(x)}{\frac{1}{x^2}} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to 0} \frac{\frac{1}{x}}{\frac{-2}{x^3}} = \lim_{x\to 0} \frac{-x^3}{2x} \\[5pt] &= \lim_{x\to 0} \frac{-x^2}{2} = 0 \end{align}$$




13. $$\lim_{x\to 0^+} x^{\text{sin}(x)}$$

    Solution

    $$\lim_{x\to 0^+} \, x^{\text{sin}(x)}$$

    This one is do-able, but we have to use some trickery. Instead, find the limit of:

    $$\lim_{x\to 0^+} \, ln\left( x^{\text{sin}(x)} \right) = \lim_{x\to 0^+} \, \text{sin}(x) ln(x)$$

    Using the properties of limits, we'll be able to take the result of this as an exponent of e to get the final answer. Now re-express as:

    $$ \begin{align} \lim_{x\to 0^+} \, \frac{ln(x)}{\frac{1}{\text{sin}(x)}} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to 0^+} \frac{\frac{1}{x}}{\frac{\text{cos}(x)}{\text{sin}^2(x)}} \\[5pt] &= \lim_{x\to 0^+} \frac{\text{sin}^2(x)}{x \text{cos}(x)} \; \longrightarrow \; \left( \frac{0}{0} \right) \\[5pt] &= \lim_{x\to 0^+} \frac{2 \text{sin}(x) \text{cos}(x)}{\text{cos}(x) - x \text{sin}(x)} = \frac{0}{1} = 0 \end{align}$$

    Now we actually found the log (ln) of the limit, so we have to undo that:

    $$\lim_{x\to 0^+} \, x^{\text{sin}(x)} = e^0 = 1$$




14. $$\lim_{x\to\infty} \frac{\sqrt{1 + x^2}}{x}$$

    Solution

    This is a fascinating limit. A first look, particularly in the context of a L'Hopital's rule section, is that it's an ∞/∞ limit, so we can apply L'Hopital's rule like this:

    $$ \begin{align} \lim_{x\to\infty} \, \frac{\sqrt{1 + x^2}}{x} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{\frac{1}{\cancel{2}}(1 + x^2)^{-1/2}\cancel{2}x}{1} \\[5pt] &= \lim_{x\to\infty} \frac{x}{\sqrt{1 + x^2}} \end{align}$$

    Now this is just the reciprocal of our original limit, and a little thought will convince you that if we apply L'Hopital's rule again, we'll just get back to the original limit; it's a never-ending loop. So how to solve it? Using the properties of limits, try squaring the whole limit:

    $$\left( \lim_{x\to \infty} \frac{\sqrt{1 + x^2}}{x} \right)^2 = \lim_{x\to \infty} \frac{1 + x^2}{x^2} = 1.$$

    Now if we just take the square root of that result, the solution is 1.




15. $$\lim_{x\to\infty} (e^{-x})^{1/x}$$

    Solution

    $$\lim_{x\to\infty} \, (e^{-x})^{\frac{1}{x}} = \lim_{x\to\infty} \left( \frac{1}{e^x} \right)^{\frac{1}{x}}$$

    This is a $0^0$ limit. We can solve it by taking the limit of the natural log of this function, remembering that we'll have to raise our result as a power of e to finish:

    $$ \begin{align} &\lim_{x\to\infty} \, ln\left[ (e^{-x})^{\frac{1}{x}} \right] \\[5pt] &= \lim_{x\to\infty} \, \frac{1}{x} \, ln(e^{-x}) = \lim_{x\to\infty}\, \frac{-x}{x} = -1 \end{align}$$

    Now raise that result as a power of e:

    $$e^{-1} = \frac{1}{e}$$




16. $$\lim_{x\to\infty} \frac{e^x}{x}$$

    Solution

    $$ \begin{align} \lim_{x\to\infty} \, \frac{e^x}{x} \; &\longrightarrow \; \left( \frac{\infty}{\infty} \right) \\[5pt] &= \lim_{x\to\infty} \frac{e^x}{1} \; \longrightarrow \; \infty \end{align}$$

    This makes sense. The exponential function in the numerator grows faster than the polynomial function in the denominator.