Introduction to Limits and L'Hôpital's Rule

1. Intuitive Idea of Limits

A limit is the value that a function approaches as the input (or variable) approaches some value. It's important to understand that we are interested in the value of the function near a point, not necessarily at the point itself.

Consider the function $f(x) = \frac{x^2 - 1}{x - 1}$. This function is not defined at $x=1$ because it leads to the form $\frac{0}{0}$. However, we can see what value $f(x)$ gets close to as $x$ gets close to 1.

For $x \neq 1$, we can simplify the function: $f(x) = \frac{(x-1)(x+1)}{x-1} = x+1$.

Now, as $x$ approaches 1 (from values slightly less than 1 or slightly more than 1), the value of $f(x)$ approaches $1+1=2$. We write this as: $$\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2$$

2. Left-Hand Limit (LHL) and Right-Hand Limit (RHL)

To have a well-defined limit at a point, the function must approach the same value from both the left and the right side.

• Right-Hand Limit (RHL): The value a function approaches as $x$ approaches a point 'a' from the right side (values greater than 'a'). It is denoted as $\lim_{x \to a^+} f(x)$. $$RHL = \lim_{h \to 0} f(a+h)$$

• Left-Hand Limit (LHL): The value a function approaches as $x$ approaches a point 'a' from the left side (values smaller than 'a'). It is denoted as $\lim_{x \to a^-} f(x)$. $$LHL = \lim_{h \to 0} f(a-h)$$

Existence of a Limit: The limit of a function $f(x)$ as $x \to a$ exists if and only if both the Left-Hand Limit and the Right-Hand Limit exist and are equal. $$\lim_{x \to a} f(x) \text{ exists} \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$ where L is a finite value.

3. Indeterminate Forms

When direct substitution of the limit point into the function results in one of the following forms, we call it an indeterminate form. This means the limit cannot be determined by simple substitution and requires further manipulation.

The seven indeterminate forms are:

$$\frac{0}{0}, \quad \frac{\infty}{\infty}, \quad 0 \times \infty, \quad \infty - \infty, \quad 0^0, \quad \infty^0, \quad 1^{\infty}$$

4.Methods for Evaluating Algebraic Limits

1. Direct Substitution: Always the first step. If it gives a finite value, that is the limit.
2. Factorization: Used for polynomial functions in a $\frac{0}{0}$ form. Factor the numerator and denominator to cancel the common factor causing the indeterminacy.
3. Rationalization: Used for expressions involving square roots, typically in a $\frac{0}{0}$ or $\infty - \infty$ form.

5. L'Hôpital's Rule (for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ forms)

L'Hôpital's Rule is a powerful method for evaluating limits of indeterminate forms of the type $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Statement: Let $f(x)$ and $g(x)$ be two functions differentiable in the neighbourhood of a point 'a' (except possibly at 'a').

If $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (the $\frac{0}{0}$ form)

OR

If $\lim_{x \to a} f(x) = \pm \infty$ and $\lim_{x \to a} g(x) = \pm \infty$ (the $\frac{\infty}{\infty}$ form)

Then, provided the limit on the right exists: $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$$

Important Notes:

• The rule can be applied successively if the new limit is also an indeterminate form.
• Always check that the limit is in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form before applying the rule. Applying it to other forms will lead to incorrect results.
• The rule is also applicable for one-sided limits ($x \to a^+$ or $x \to a^-$) and for limits where $x \to \infty$ or $x \to -\infty$.

Solved Examples

Example 1: Applying L'Hôpital's Rule (0/0 form)

Question: Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$.

Solution:

1. Check Form: As $x \to 0$, $\sin x \to 0$ and $x \to 0$. This is the $\frac{0}{0}$ indeterminate form.
2. Apply L'Hôpital's Rule: We differentiate the numerator and the denominator with respect to $x$. Let $f(x) = \sin x \implies f'(x) = \cos x$. Let $g(x) = x \implies g'(x) = 1$.
3. Evaluate New Limit: $$\lim_{x \to 0} \frac{\cos x}{1} = \frac{\cos 0}{1} = \frac{1}{1} = 1$$ Therefore, $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Example 2: Applying L'Hôpital's Rule Twice

Question: Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Solution:

1. Check Form: As $x \to 0$, $1 - \cos x \to 1 - 1 = 0$ and $x^2 \to 0$. This is the $\frac{0}{0}$ form.
2. Apply L'Hôpital's Rule (First Time): $$\lim_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x^2)} = \lim_{x \to 0} \frac{\sin x}{2x}$$
3. Check Form Again: The new limit is still of the form $\frac{0}{0}$ as $x \to 0$.
4. Apply L'Hôpital's Rule (Second Time): $$\lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{\cos x}{2}$$
5. Evaluate Final Limit: $$\frac{\cos 0}{2} = \frac{1}{2}$$ Therefore, $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.

Example 3: Applying L'Hôpital's Rule (∞/∞ form)

Question: Evaluate $\lim_{x \to \infty} \frac{\ln x}{x}$.

Solution:

1. Check Form: As $x \to \infty$, $\ln x \to \infty$ and $x \to \infty$. This is the $\frac{\infty}{\infty}$ indeterminate form.
2. Apply L'Hôpital's Rule: $$\lim_{x \to \infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x)} = \lim_{x \to \infty} \frac{1/x}{1} = \lim_{x \to \infty} \frac{1}{x}$$
3. Evaluate Final Limit: $$\lim_{x \to \infty} \frac{1}{x} = 0$$ Therefore, $\lim_{x \to \infty} \frac{\ln x}{x} = 0$.