Section 4-4: Mastering L'Hospital's Rule

Welcome to a deep dive into L'Hospital's Rule! This powerful tool allows us to evaluate limits that initially present in indeterminate forms. Don't be intimidated; with a clear understanding of the conditions and steps, you'll be solving these problems with confidence.

What is L'Hospital's Rule?

L'Hospital's Rule is applicable when we encounter limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Specifically, suppose we have two differentiable functions, $f(x)$ and $g(x)$, where $g'(x) \neq 0$ on an open interval $I$ containing $a$ (except possibly at $a$). If:

  • $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, OR
  • $\lim_{x \to a} f(x) = \pm \infty$ and $\lim_{x \to a} g(x) = \pm \infty$

Then, L'Hospital's Rule states:

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

provided the limit on the right side exists (or is $\infty$ or $-\infty$).

Examples to Illuminate the Path

Let's work through some examples to see L'Hospital's Rule in action. Remember to always check if the limit results in an indeterminate form before applying the rule!

Example 1: $\frac{0}{0}$ Form

Find the limit:

$$ \lim_{x \to 1} \frac{\ln x}{x - 1} $$

As $x \to 1$, $\ln x \to 0$ and $x - 1 \to 0$, so we have the indeterminate form $\frac{0}{0}$. Applying L'Hospital's Rule:

Let $f(x) = \ln x$ and $g(x) = x - 1$. Then $f'(x) = \frac{1}{x}$ and $g'(x) = 1$.

$$ \lim_{x \to 1} \frac{\ln x}{x - 1} = \lim_{x \to 1} \frac{\frac{1}{x}}{1} = \lim_{x \to 1} \frac{1}{x} = 1 $$

Example 2: $\frac{\infty}{\infty}$ Form

Calculate the limit:

$$ \lim_{x \to \infty} \frac{e^x}{x^2} $$

As $x \to \infty$, both $e^x$ and $x^2$ approach $\infty$, giving us the indeterminate form $\frac{\infty}{\infty}$. Applying L'Hospital's Rule:

Let $f(x) = e^x$ and $g(x) = x^2$. Then $f'(x) = e^x$ and $g'(x) = 2x$. Applying the rule once:

$$ \lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{e^x}{2x} $$

We still have the form $\frac{\infty}{\infty}$, so we apply L'Hospital's Rule again:

$f'(x) = e^x$ and $g'(x) = 2$.

$$ \lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{e^x}{2} = \infty $$

Example 3: Limits approaching 0

Calculate the limit:

$$ \lim_{x \to 0^+} x \ln x $$

Rewrite this as

$$ \lim_{x \to 0^+} \frac{\ln x}{1/x} $$

As $x \to 0^+$, both $\ln x$ approaches $-\infty$ and $1/x$ approaches $\infty$, giving us the indeterminate form $\frac{-\infty}{\infty}$. Applying L'Hospital's Rule:

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

Important Considerations

  • Always check for an indeterminate form. L'Hospital's Rule only applies to forms like $\frac{0}{0}$ and $\frac{\infty}{\infty}$.
  • Differentiate correctly. Double-check your derivatives!
  • Apply the rule repeatedly if necessary. As seen in Example 2, you might need to apply L'Hospital's Rule multiple times.

Keep practicing, and you'll become proficient in using L'Hospital's Rule to conquer challenging limits. Happy calculating!