4.7 Using L'Hôpital's Rule for Determining Limits in Indeterminate Forms

Welcome to the last key topic of unit 4! 🥳

You may recall from Unit 1 that sometimes the limits of functions evaluate to $\pm\frac{\infty}{\infty}$ or $\frac{0}{0}$ which are indeterminate forms. Now, instead of looking for another way to manipulate the equation to try and find an answer that is not in indeterminate form, we can use our knowledge of derivatives to help us! Specifically, we can use L’Hopital’s Rule. ⭐

📏 L’Hopital’s Rule

L’Hopital’s Rule states that if $\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{0}{0}$ or $\lim_{x\to a}\frac{f(x)}{g(x)}=\pm\frac{\infty}{\infty}$,

Basically, the rule states that we can evaluate the limits of indeterminate forms using their derivatives!

✏️ L’Hopital’s Rule: Walkthrough

Let’s try a practice problem together! Evaluate the following limit.

Plugging $x=\frac{\pi}{2}$ into $\frac{\cos(x)}{x-\frac{\pi}{2}}$ results in the indeterminate form $\frac{0}{0}$. This signals to us that we should use L’Hopital’s Rule.

When completing a free-response question on the AP exam, it is important to show the limits of f(x), the numerator, and g(x), the denominator, are separately equal to the needed parameters. Let’s go ahead and do that now!

Since $\lim_{x \to \frac{\pi}{2}}{\cos(x)}=0$ and $\lim_{x \to \frac{\pi}{2}}x-\frac{\pi}{2}=0$, L’Hopital’s Rule can be applied. Be sure to write this statement out before actually applying this rule.

Now, we can take the derivatives and get into L’Hopital’s Rule.

In conclusion, we know that this limit…

Great work! Doesn’t L’Hopital’s Rule save so much time? 🙃

📝 Practicing L’Hopital’s Rule

Here are some problems you can practice applying L’Hopital’s Rule on!

❓L’Hopital’s Rule: Practice Problems

Evaluate the following limits. Imagine these are free-response questions and you have to check conditions!

Question 1: L’Hopital’s Rule

Question 2: L’Hopital’s Rule

✅ L’Hopital’s Rule: Answers and Solutions

Question 1: L’Hopital’s Rule

Plugging $x=0$ into $\frac{\tan(x)}{7x+\tan(x)}$ results in the indeterminate form $\frac{0}{0}$. Since the expression involves mixed function types, it is not possible to manipulate it algebraically in any way to find the limits. Therefore, we should use L’Hopital’s Rule.

But first, show that the limits are separately equal to 0.

Since $\lim_{x\to 0}\tan(x)=0$ and $\lim_{x\to 0}7x+\tan(x)=0$, L’Hopital’s Rule can be applied.

In conclusion…

Question 2: L’Hopital’s Rule

Plugging $x=\infty$ into $\frac{3x^2-8}{7x^2+21}$ results in the indeterminate form $\frac{\infty}{\infty}$. Therefore, we should use L’Hopital’s Rule.

Since $\lim_{x\to \infty}3x^2-8=\infty$ and $\lim_{x\to \infty}7x^2+21=\infty$, L’Hopital’s Rule can be applied.

Therefore…

💫 Closing

Remember, L'Hopital's Rule provides a powerful method for evaluating limits involving indeterminate forms. By taking derivatives of the numerator and denominator, it helps simplify complex limits, leading to easier evaluation.

Happy calculus studying! 📚