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Understanding Limits, L’Hôpital’s Rule, and Discontinuities

In this guide we’ll walk through classic calculus limit problems, the types of discontinuities you’ll see, and how tools like factoring, L’Hôpital’s Rule, and the Squeeze Theorem help.

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1. Removable Discontinuity

We want:

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$

Direct substitution:

$$\frac{(2)^2 - 4}{2 - 2} = \frac{0}{0}$$

This is an indeterminate form.

Factor the numerator:

$$x^2 - 4 = (x - 2)(x + 2)$$

For $x \neq 2$ :

$$\frac{x^2 - 4}{x - 2} = x + 2$$

The graph looks like $y = x + 2$ except for a hole at $x = 2$ .

Limit:

$$\lim_{x \to 2} (x + 2) = 4$$

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1. L’Hôpital’s Rule

When to use: Direct substitution gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

If $f$ and $g$ are differentiable near $a$ , $g'(x) \neq 0$ near $a$ , and the derivative quotient limit exists:

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

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Example:

$$\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$$

Differentiate top and bottom:

$$f'(x) = 2x, \quad g'(x) = 1$$

Then:

$$\lim_{x \to 2} \frac{2x}{1} = 4$$

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Example with $\infty/\infty$ :

$$\lim_{x \to \infty} \frac{3x^2 + 5x}{7x^2 - 4}$$

Differentiate:

$$f'(x) = 6x + 5, \quad g'(x) = 14x$$

So:

$$\lim_{x \to \infty} \frac{6 + \frac{5}{x}}{14} = \frac{3}{7}$$

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1. Classic Limit — $\sin x / x$

$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

By L’Hôpital:

$$\lim_{x \to 0} \frac{\cos x}{1} = 1$$

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1. Infinite Discontinuity

For $\frac{1}{x}$ :

As $x \to 0^+$ : $\frac{1}{x} \to +\infty$

As $x \to 0^-$ : $\frac{1}{x} \to -\infty$

Two-sided limit does not exist.

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1. Jump Discontinuity

$$f(x) = \begin{cases} 1 & x < 0 \ 2 & x \ge 0 \end{cases}$$

The function jumps from 1 to 2 at $x = 0$ .

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1. Limit Does Not Exist — $\sin(1/x)$

As $x \to 0$ , $1/x \to \pm\infty$ and $\sin(1/x)$ oscillates infinitely between −1 and 1.
No single limit value exists.

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1. Piecewise Jump Discontinuity

$$f(x) = \begin{cases} x^2 - 1 & x < 1 \ 2 & x = 1 \ 3 - x & x > 1 \end{cases}$$

LHL at $x=1$ is $0$

RHL at $x=1$ is $2$

Since LHL ≠ RHL, there’s a jump discontinuity.

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1. Squeeze Theorem Example

Let:

$$f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right), & x \neq 0 \ 0, & x = 0 \end{cases}$$

We know:

$$-1 \le \sin\left(\frac{1}{x}\right) \le 1$$

Multiply by $|x|$ :

$$-|x| \le x \sin\left(\frac{1}{x}\right) \le |x|$$

Both bounds → $0$ as $x \to 0$ .
By the Squeeze Theorem:

$$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$$

✅ Continuous at $x = 0$ .

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