L'Hôpital's Rule and Indeterminate Forms
When calculating some limits, it may lead to indeterminate forms which means evaluating the limit directly is not possible. Some of these forms can be worked around by rearranging, and for some we can use the rule of l'hopital.
Indeterminate Forms
The two main indeterminate forms that we can use l'hopital on are:
- $\frac{0}{0}$: This occurs when both the numerator and denominator approach zero as $x$ approaches a certain value.
- $\frac{\infty}{\infty}$: This happens when both the numerator and denominator grow without bound as $x$ approaches a certain value or infinity.
Other indeterminate forms exist, such as $0 \cdot \infty$, $\infty - \infty$, $0^0$, $1^\infty$, and $\infty^0$, but not all can be used with l'hopital
But some forms can be rearranged to match the ones listed above!
L'Hôpital's Rule
L'Hôpital's Rule states that for functions $f(x)$ and $g(x)$ that are differentiable near a point $a$, if the limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ results in an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$
provided that the limit of the ratio of derivatives exists or is infinite.
L'Hôpital's Rule can also be applied when $a$ is $\infty$ or $-\infty$.
Applying L'Hôpital's Rule
To apply L'Hôpital's Rule:
- Verify that the limit results in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$).
- Differentiate both the numerator and denominator separately.
- Take the limit of the ratio of these derivatives.
- If the result is still indeterminate, repeat steps 2 and 3 until a determinate form is reached.
Let's evaluate $\lim_{x \to 0} \frac{\sin x}{x}$:
- As $x \to 0$, both $\sin x \to 0$ and $x \to 0$, giving the indeterminate form $\frac{0}{0}$.
- Apply L'Hôpital's Rule: $$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{(\sin x)'}{(x)'} = \lim_{x \to 0} \frac{\cos x}{1} = 1$$
Thus, $\lim_{x \to 0} \frac{\sin x}{x} = 1$.
Sometimes, a single application of L'Hôpital's Rule may result in another indeterminate form. In such cases, the rule can be applied repeatedly until a determinate form is reached.
Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$:
- As $x \to 0$, both $1 - \cos x \to 0$ and $x^2 \to 0$, giving $\frac{0}{0}$.
- First application: $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x}$$
- This is still $\frac{0}{0}$, so apply L'Hôpital's Rule again: $$\lim_{x \to 0} \frac{\sin x}{2x} = \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$$
Therefore, $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.
Maclaurin Series Method
An alternative to L'Hôpital's Rule for evaluating limits is using Maclaurin series expansions. This method involves expanding functions into their Maclaurin series and then evaluating the limit (more in chapter 5.19)
The Maclaurin series for a function $f(x)$ is given by:
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$
Let's revisit $\lim_{x \to 0} \frac{\sin x}{x}$ using Maclaurin series:
- The Maclaurin series for $\sin x$ is: $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$
- Substituting this into the limit: $$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots}{x}$$
- Cancelling $x$ from numerator and denominator: $$\lim_{x \to 0} (1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots) = 1$$
This confirms our earlier result using L'Hôpital's Rule.
You can only use the Maclaurin Series expansion for limits that tend to $0$, otherwise you have to use the Taylor series which is not part of the course!
Connection to Horizontal Asymptotes
The techniques of evaluating limits, particularly as $x \to \infty$, are closely related to finding horizontal asymptotes of functions.
A horizontal asymptote for a function $y = f(x)$ is a line $y = L$ where:
$$\lim_{x \to \infty} f(x) = L$$
For rational functions $f(x) = \frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials:
- If degree of $P< $ degree of $Q$, the horizontal asymptote is $y = 0$.
- If degree of $P = $ degree of $Q$, the horizontal asymptote is $y = $ ratio of leading coefficients.
- If degree of $P > $ degree of $Q$, there is no horizontal asymptote (the function has a slant asymptote instead).
For $f(x) = \frac{2x^2 + 3x - 1}{x^2 - 4}$, find the horizontal asymptote:
$$\lim_{x \to \infty} \frac{2x^2 + 3x - 1}{x^2 - 4} = \lim_{x \to \infty} \frac{2 + \frac{3}{x} - \frac{1}{x^2}}{1 - \frac{4}{x^2}} = 2$$
Therefore, the horizontal asymptote is $y = 2$.
Students often forget to check if L'Hôpital's Rule is applicable before applying it. Always verify that you have an indeterminate form first!
By mastering these techniques, students can tackle a wide range of limit problems and gain deeper insights into the behavior of functions, particularly near critical points and as they approach infinity.
