When you teach a person, actually two people learn - here i am delving into calculus to show the light to my son...

1. The Problem

We have:

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

If we try direct substitution ($x = 2$):

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

This is an indeterminate form — so the limit is not obvious.

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Why it’s a Removable Discontinuity

The numerator $x^2 - 4$ factors:

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

So:

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

For $x \neq 2$, the $(x - 2)$ cancels:

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

The function behaves exactly like $y = x + 2$ everywhere except at $x = 2$.

At $x = 2$, the original expression was undefined (division by zero), so there’s a hole in the graph — that’s the removable discontinuity.

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Finding the Limit

Now that it’s simplified:

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

So:

$$\boxed{\text{The limit is } 4}$$

If we define $f(2) = 4$, the discontinuity disappears and the function becomes continuous.

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Visualizing

If you graph $y = \frac{x^2 - 4}{x - 2}$:

• It looks like a straight line $y = x + 2$

• Except there’s an open circle at $(2, 4)$
  That hole is the removable discontinuity.

Here's a visual representation of f(x).

The left top (the first one) shows the f(x) plotted.

Now let’s walk through L’Hôpital’s Theorem for limits clearly and step-by-step.

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The Problem L’Hôpital Solves

Sometimes, when evaluating a limit, direct substitution gives an indeterminate form like:

• $\frac{0}{0}$

• $\frac{\infty}{\infty}$

These forms mean you can’t just plug in the value — but the limit might still exist.

Example:

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

Direct substitution → $0/0$.

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Statement of L’Hôpital’s Rule

If:

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

2. $f$ and $g$ are differentiable near $a$ (except possibly at $a$)

3. $g'(x) \neq 0$ near $a$

Then:

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

provided the limit on the right exists or is $\pm\infty$.

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How It Works (Example)

Example:

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

Direct substitution → $0/0$.

Step 1: Differentiate numerator and denominator separately:

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

Step 2: Apply L’Hôpital’s Rule:

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

Same answer we got earlier by factoring — but L’Hôpital works even when factoring is messy.

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Another Example with $\infty/\infty$

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

Direct substitution → $\infty/\infty$.

Differentiate top and bottom:

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

Now:

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

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Notes and Common Mistakes

• Only use when you get $0/0$ or $\infty/\infty$
  (For other indeterminate forms like $0 \cdot \infty$, $1^\infty$, $\infty - \infty$, you may first need algebraic manipulation.)

• Apply the rule repeatedly if the first derivative still gives $0/0$ or $\infty/\infty$.

• Remember: Differentiate top and bottom separately, never use the quotient rule here.

2. The Problem

Alright — let’s go step-by-step for

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

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Direct Substitution Check

If $x = 0$,

$$\frac{\sin(0)}{0} = \frac{0}{0}$$

That’s an indeterminate form — so we need another method.

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

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Using L’Hôpital’s Rule

We have $0/0$ form. Differentiate top and bottom:

$$\underset{x\to 0}{\lim }\frac{\cos x}{1}=\cos 0=1$$

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Final Answer:

                                                                        1

Visualization

See the plot at left column second from top.

3 . The problem

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

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First observation

If $x = 0$, the expression is undefined because we can’t divide by zero.
So we check left-hand limit (LHL) and right-hand limit (RHL) separately.

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Right-hand limit ( $x \to 0^+$ )

Take $x$ as small positive numbers:

$x$	$1/x$
0.1	10
0.01	100
0.001	1000

As $x \to 0^+$, $1/x \to +\infty$.

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Left-hand limit ( $x \to 0^-$ )

Take $x$ as small negative numbers:

$x$	$1/x$
-0.1	-10
-0.01	-100
-0.001	-1000

As $x \to 0^-$, $1/x \to -\infty$.

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Conclusion

The two one-sided limits are not equal:

$$\lim_{x \to 0^+} \frac{1}{x} = +\infty$$ $$\lim_{x \to 0^-} \frac{1}{x} = -\infty$$

Since they differ, the two-sided limit does not exist.

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📌 Important: This is an example of an infinite discontinuity.

This is the third graph in the image attached earlier.

4. The problem - Jump Discontinuity

$$f(x)=\{\begin{array}{ll}1& x<0\\ 2& x\ge \mathrm{0<span\; style="text-align:\; left;"></span>}\end{array}$$

Visualization:

A step function shows a sudden jump — perfect for illustrating discontinuity.

The topmost in the right column of the image attached earlier

4. The Problem

The limit of

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

is a classic example of a limit that does not exist.

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Why it doesn’t exist

As $x \to 0$,

• $\frac{1}{x} \to +\infty$ when $x \to 0^+$

• $\frac{1}{x} \to -\infty$ when $x \to 0^-$

The sine function $\sin(t)$ oscillates between -1 and +1 infinitely often as $t \to \pm \infty$.

So near $x=0$, $\sin\left(\frac{1}{x}\right)$ oscillates wildly — no single number is approached.

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Formal reasoning

We can try to take the right-hand limit:

Pick sequences:

$$x_n = \frac{1}{2\pi n} \quad \Rightarrow \quad \sin\left(\frac{1}{x_n}\right) = \sin(2\pi n) = 0$$

and

$$y_n = \frac{1}{2\pi n + \pi/2} \quad \Rightarrow \quad \sin\left(\frac{1}{y_n}\right) = \sin\left(2\pi n + \frac{\pi}{2}\right) = 1$$

Both $x_n \to 0$ and $y_n \to 0$, but the function values approach different limits (0 and 1).
Hence, the limit does not exist.

5. The problem

Alright, let’s check if your function is continuous at $x = 1$.

Given:

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

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Step 1 – Left-hand limit (as $x \to 1^-$)

For $x < 1$, $f(x) = x^2 - 1$

$$\lim_{x \to 1^-} f(x) = 1^2 - 1 = 0$$

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Step 2 – Right-hand limit (as $x \to 1^+$)

For $x > 1$, $f(x) = 3 - x$

$$\lim_{x \to 1^+} f(x) = 3 - 1 = 2$$

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Step 3 – Compare limits

• Left-hand limit = 0

• Right-hand limit = 2
  Since $0 \neq 2$, the limit at $x = 1$ does not exist.

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Step 4 – Conclusion

Continuity at $x = 1$ requires:

$$\lim_{x \to 1} f(x) = f(1)$$

But here:

• The two one-sided limits are not equal

• So the function has a jump discontinuity at $x = 1$

Final Answer: ❌ Not continuous at $x = 1$.

Visualization

The bottom of the left column shows the graphical output of the function

6. The Problem

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