Welcome back to Professor Baker's Math Class! In Section 4-4, we are upgrading our calculus toolkit with one of the most useful theorems in the course: L'Hospital's Rule. Back in Chapter 2, when we encountered limits that resulted in $\frac{0}{0}$, we had to rely on algebraic gymnastics like factoring, rationalizing, or finding common denominators. Today, we learn a faster, more elegant way to handle these indeterminate forms using derivatives.

What is an Indeterminate Form?

An indeterminate form occurs when direct substitution into a limit gives us a result that doesn't have a defined value. The two most common "classic" forms are:

  • $\frac{0}{0}$
  • $\frac{\infty}{\infty}$ (or $\frac{-\infty}{\infty}$, etc.)

Crucial Step: Always plug in your $x$-value first! As seen in our notes (referencing the $\frac{\sin x}{1-\cos x}$ example), sometimes a limit is defined immediately (like $\frac{0}{2} = 0$), and applying L'Hospital's Rule blindly would give you the wrong answer.

L'Hospital's Rule Defined

Suppose that $\lim_{x \to a} \frac{f(x)}{g(x)}$ results in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$). L'Hospital's Rule states that:

$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Provided the limit on the right side exists. This means we can take the derivative of the numerator and the derivative of the denominator separately and then try the limit again.

Warning: Do not use the Quotient Rule here! We are differentiating the top and bottom independently.

Example: The Basic Application

Let's look at a classic example from the notes:

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

Direct substitution gives $\frac{1-1}{1-1} = \frac{0}{0}$. Applying L'Hospital's Rule:

  1. $f(x) = x^2-x \rightarrow f'(x) = 2x-1$
  2. $g(x) = x^2-1 \rightarrow g'(x) = 2x$

Now, evaluate the new limit:

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

Repeated Application

Sometimes, applying the rule once isn't enough. Consider the limit as $x \to \infty$ for $\frac{e^x}{x^2}$.

  • First check: $\frac{\infty}{\infty}$. Apply L'Hospital's Rule $\rightarrow \frac{e^x}{2x}$.
  • Second check: Still $\frac{\infty}{\infty}$. Apply it again! $\rightarrow \frac{e^x}{2}$.
  • Final result: As $x \to \infty$, $\frac{e^x}{2} \to \infty$.

Advanced Indeterminate Forms

Not all indeterminate forms look like fractions. We also encounter products ($0 \cdot \infty$), differences ($\infty - \infty$), and powers ($0^0, 1^\infty, \infty^0$). The strategy here is to manipulate the function into a fraction so we can use L'Hospital's Rule.

1. Indeterminate Products ($0 \cdot \infty$)

If you have $\lim f(x)g(x)$, you can rewrite it as $\frac{f(x)}{1/g(x)}$.

Example: $\lim_{x \to 0^+} x \ln x$. This is $0 \cdot (-\infty)$.
Rewrite it as $\frac{\ln x}{x^{-1}}$. Now it is in the form $\frac{-\infty}{\infty}$. Differentiate top and bottom:

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

2. Indeterminate Powers ($x^x$)

For limits like $\lim_{x \to 0^+} x^x$, we use natural logs to bring the exponent down.

  1. Let $y = x^x$
  2. Take $\ln$ of both sides: $\ln y = x \ln x$
  3. Find the limit of $\ln y$ (which we just found above is 0).
  4. Since $\ln y \to 0$, then $y \to e^0 = 1$.

Keep practicing these transformations. L'Hospital's Rule is a powerful ally, turning impossible algebra problems into manageable derivative problems!