Limits: L'Hospital's rule

We know that when we consider such limits as $\lim\limits_{x \to a} \dfrac{f(x)}{g(x)}$, we can get the $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ indeterminate forms and can't interpret this result. Fortunately, there is a rule we can apply in such cases: it is called the L'Hospital's rule.

What is the L'Hospital's rule?

Suppose we need to find $\lim\limits_{x \to a} \dfrac{f(x)}{g(x)}$. If both functions $f(x)$ and $g(x)$ are differentiable and both of them approach $0$ as $x\to a$:

$$\lim\limits_{x \to a}f(x) = 0 \ \ \lim\limits_{x \to a}g(x) = 0$$our limit can be evaluated if we take the derivative of each function:

$$\lim\limits_{x \to a} \dfrac{f(x)}{g(x)} = \lim\limits_{x \to a} \dfrac{f'(x)}{g'(x)}$$Consider the example:

$$\lim\limits_{x \to 0} \dfrac{\sin x}{x}$$Since $\sin x \to 0$ when $x \to 0$, then we have the $\dfrac{0}{0}$ indeterminate form and we can apply the L'Hospital's rule:

$$\lim\limits_{x \to 0} \dfrac{\sin x}{x} = \lim\limits_{x \to 0} \dfrac{(\sin x)'}{x'} = \lim\limits_{x \to 0} \dfrac{\cos x}{1} = \lim\limits_{x \to 0}\cos x = 1$$In addition to the $\dfrac{0}{0}$ indeterminate form, the L'Hospital's rule is applicable even in case of the $\dfrac{\infty}{\infty}$ indeterminate form. Suppose we need to find the limit

$$\lim\limits_{x \to \infty} \dfrac{3e^x}{x^2}$$This limit gives us the $\dfrac{\infty}{\infty}$ indeterminate form, because if $x \to \infty$, then $3e^x \to \infty$ and $x^2 \to \infty$. Let's use the L'Hospital's rule:

$$\lim\limits_{x \to \infty} \dfrac{3e^x}{x^2} = \lim\limits_{x \to \infty} \dfrac{(3e^x)'}{(x^2)'} = \lim\limits_{x \to \infty} \dfrac{3e^x}{2x}$$Now we still have the $\dfrac{\infty}{\infty}$ indeterminate form, so let's take the derivative again. If the limit of the quotient is the same as the limit of the quotient of the derivatives, then our rule will work, no matter how many times we take the derivative.

$$\lim\limits_{x \to \infty} \dfrac{(3e^x)'}{(2x)'} = \lim\limits_{x \to \infty} \dfrac{3e^x}{2} = \dfrac{3}{2} \lim\limits_{x \to \infty} e^x = \infty$$So, $$\lim\limits_{x \to \infty} \dfrac{3e^x}{x^2} = \infty$$Notice, that the L'Hospital's rule only applies to limits of $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ indeterminate forms.

Another form of the L'Hospital's rule

Consider the limit

$$\lim\limits_{x \to a} f(x) \cdot g(x)$$where $\lim\limits_{x \to a} f(x) = 0$ and $\lim\limits_{x \to a} g(x) = \infty$, so we have the $0 \cdot \infty$ indeterminate form. We can't directly determine what is going on here. But we can rewrite our limit in another way:

$$\lim\limits_{x \to a}f(x) \cdot g(x) = \lim\limits_{x \to a} \dfrac{f(x)}{\frac{1}{g(x)}}$$or

$$\lim\limits_{x \to a}f(x) \cdot g(x) = \lim\limits_{x \to a} \dfrac{g(x)}{\frac{1}{f(x)}}$$

We got $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ indeterminate forms and now we can apply the L'Hospital's rule!

For example:

$$\lim\limits_{x \to 0+} (x^2 \cdot \ln x)$$Here we have the $0 \cdot (-\infty)$ indeterminate form, because

$$\lim\limits_{x \to 0+} x^2 = 0$$

and

$$\lim \limits_{x \to 0+} \ln x = -\infty$$

Let's rewrite the product $x^2 \cdot \ln x$ as $\dfrac{\ln x}{\frac{1}{x^2}}$and apply L'Hospital's rule:

$$\lim\limits_{x \to 0+} (x^2 \cdot \ln x) = \lim\limits_{x \to 0+} \dfrac{\ln x}{\frac{1}{x^2}} = \lim\limits_{x \to 0+} \dfrac{(\ln x)'}{(\frac{1}{x^2})'} = \\ \ \\ = \lim\limits_{x \to 0+} \dfrac{\frac{1}{x}}{\frac{-2}{x^3}} = - \lim\limits_{x \to 0+} \dfrac{x^2}{2} = 0$$

What about indeterminate powers?

Suppose we have $$\lim\limits_{x \to a} f(x)^{g(x)}$$and this is a $0^0$, $\infty^0$ or $1^\infty$ indeterminate form. In such cases we can set the expression $f(x)^{g(x)}$ to $y$ and take the natural logarithm of both sides:

$$y = f(x)^{g(x)} \\ \ln y = \ln f(x)^{g(x)} \\ \ln y = g(x) \cdot \ln f(x) \\ y = e^{g(x) \cdot \ln f(x)}$$So,

$$\lim\limits_{x \to a} f(x)^{g(x)} = \lim\limits_{x \to a} e^{g(x) \cdot \ln f(x)}$$Now we may be able to evaluate our limit.

Consider the example:

$$\lim\limits_{x \to 0+} x^x$$

We convert this expression:

$$y = x^x \\ \ln y = \ln x^x \\ \ln y = x \cdot \ln x \\ y = e^{x \cdot \ln x}$$Then

$$\lim\limits_{x \to 0+} x^x = \lim\limits_{x \to 0+} e^{x \cdot \ln x}$$Using the method described in the previous section, we obtain

$$\lim\limits_{x \to 0+} x \cdot \ln x = 0$$

So,

$$\lim\limits_{x \to 0+} x^x = \lim\limits_{x \to 0+} e^{x \cdot \ln x} = \lim\limits_{x \to 0+} e^0 = 1$$

Conclusion

In this topic, we have learned about the L'Hospital's rule. When we want to use this rule, we must remember that the L'Hospital's rule only applies to limits with $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ indeterminate forms. This should always be checked, otherwise you may get the wrong answer.