Introduction

L'Hôpital's Rule is an important technique that makes our lives easier when dealing with certain forms of tricky limits — those that evaluate to indeterminate form.

Identifying Indeterminate Form

Indeterminate Form

Indeterminate form arises when evaluating a limit leads to an expression that cannot be truly evaluated. There are several different kinds:

Quotient Forms are the most common:

• $\dfrac{0}{0}$
• $\dfrac{\infty}{\infty}$

Other Forms include:

• Product Form: $0 \cdot \infty$ (e.g. $\lim_{x \to 0^+} x \ln x$)
• Difference Form: $\infty - \infty$ (e.g. $\lim_{x \to 0^+} \dfrac{1}{x^2} - \dfrac{1}{\tan x}$)
• Power Form: $\infty^0$, $0^0$, or $1^\infty$ (e.g. $\lim_{x \to \infty} x^{1/x}$)

Getting to Quotient Form

L'Hôpital's Rule requires the limit to be in quotient form. Non-quotient indeterminate forms must first be algebraically rearranged into a fraction — the examples below illustrate how.

Applying L'Hôpital's Rule

Once a limit is confirmed to be in indeterminate quotient form, L'Hôpital's Rule states:

If $\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)}$ evaluates to $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then:

If the result is still indeterminate, the rule can be applied repeatedly.

Worked Examples

Example 1: Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin(x)}{x}$.

Solution: Both $\sin(x)$ and $x$ approach $0$ as $x \to 0$, giving the form $\dfrac{0}{0}$. Setting $f(x) = \sin(x)$ and $g(x) = x$, we apply L'Hôpital's Rule:

$\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1} = \boxed{1}$

Example 2: Evaluate $\displaystyle\lim_{x \to \infty} \frac{3x + 5}{2x + 1}$.

Solution: As $x \to \infty$, both numerator and denominator grow without bound, giving $\dfrac{\infty}{\infty}$. Applying L'Hôpital's Rule:

$\lim_{x \to \infty} \frac{3x + 5}{2x + 1} = \lim_{x \to \infty} \frac{3}{2} = \boxed{\frac{3}{2}}$

Example 3: Evaluate $\displaystyle\lim_{x \to 0^+} x \ln x$.

Solution: As $x \to 0^+$, we have $x \to 0$ and $\ln x \to -\infty$, giving the form $0 \cdot \infty$. Rewrite as a quotient by dividing instead of multiplying:

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

Now apply L'Hôpital's Rule (this is $\dfrac{\infty}{\infty}$):

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

Example 4: Evaluate $\displaystyle\lim_{x \to 0^+} \left(\frac{1}{x^2} - \frac{1}{\tan x}\right)$.

Solution: As $x \to 0^+$, both terms approach $\infty$, giving the form $\infty - \infty$. Combine into a single fraction by cross-multiplying:

$\lim_{x \to 0^+} \frac{\tan(x) - x^2}{x^2 \tan(x)}$

Apply L'Hôpital's Rule:

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

As $x \to 0^+$: the numerator approaches $\sec^2(0) = 1$, while the denominator approaches $0$. Hence the limit is $\boxed{\infty}$.

Example 5: Evaluate $\displaystyle\lim_{x \to \infty} x^{1/x}$.

Solution: This has the form $\infty^0$. Take the natural log to convert to a quotient:

$\ln\!\left(x^{1/x}\right) = \frac{\ln x}{x}$

Using the continuity of $\ln$:

$\ln\!\left(\lim_{x \to \infty} x^{1/x}\right) = \lim_{x \to \infty} \frac{\ln x}{x}$

Apply L'Hôpital's Rule (this is $\dfrac{\infty}{\infty}$):

$= \lim_{x \to \infty} \frac{1/x}{1} = 0$

Therefore $\ln\!\left(\displaystyle\lim_{x \to \infty} x^{1/x}\right) = 0$, which gives $\displaystyle\lim_{x \to \infty} x^{1/x} = e^0 = \boxed{1}$.

Proof

Starting with a quotient limit that evaluates to $\dfrac{0}{0}$, we know $f(a) = g(a) = 0$:

Since the limit of a quotient equals the quotient of the limits (when both exist):

The numerator and denominator are precisely the difference quotient definitions of $f'$ and $g'$.

Shortcomings

L'Hôpital's Rule only works when the limit is in indeterminate form. Misapplying it to a determinate limit yields a wrong answer.

Example (misuse): Evaluate $\displaystyle\lim_{x \to 1} \frac{x^2 + 5}{3x + 4}$.

Correct approach: Substituting $x = 1$ directly: $\dfrac{1 + 5}{3 + 4} = \dfrac{6}{7}$.

Incorrect use of L'Hôpital's Rule: Since this is not indeterminate form, applying the rule gives $\displaystyle\lim_{x \to 1} \frac{2x}{3} = \frac{2}{3}$, which is wrong. Always check for indeterminate form before applying the rule.