4.1 Limits & Continuity

Detailed Theory: Limits & Continuity

1. Introduction to Limits

1.1 What is a Limit?

The limit describes the behavior of a function as its input approaches a certain value.

Informal Definition: $\lim_{x \to a} f(x) = L$ means that as $x$ gets closer and closer to $a$, the function values $f(x)$ get closer and closer to $L$.

Example: Consider $f(x) = \frac{x^2 - 1}{x - 1}$

What happens as $x$ approaches $1$?

For $x \neq 1$: $f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1$

As $x \to 1$, $x + 1 \to 2$

So $\lim_{x \to 1} f(x) = 2$

1.2 Notation

• $\lim_{x \to a} f(x) = L$: Limit as $x$ approaches $a$

• $\lim_{x \to a^+} f(x)$: Right-hand limit ($x$ approaches from right)

• $\lim_{x \to a^-} f(x)$: Left-hand limit ($x$ approaches from left)

1.3 One-Sided Limits

Right-hand limit: $\lim_{x \to a^+} f(x) = L$ means as $x$ approaches $a$ from values greater than $a$, $f(x)$ approaches $L$.

Left-hand limit: $\lim_{x \to a^-} f(x) = L$ means as $x$ approaches $a$ from values less than $a$, $f(x)$ approaches $L$.

Important: For $\lim_{x \to a} f(x)$ to exist, both one-sided limits must exist and be equal.

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2. Formal Definition of Limit

2.1 ε-δ Definition (Precise Definition)

$\lim_{x \to a} f(x) = L$ means:

For every $\epsilon > 0$, there exists $\delta > 0$ such that:

If $0 < |x - a| < \delta$, then $|f(x) - L| < \epsilon$

In words: We can make $f(x)$ as close as we want to $L$ by taking $x$ sufficiently close to $a$ (but not equal to $a$).

2.2 Understanding the Definition

• $\epsilon$ (epsilon): How close we want $f(x)$ to be to $L$

• $\delta$ (delta): How close $x$ needs to be to $a$ to achieve that

• The definition works for ALL positive $\epsilon$, no matter how small

Example: Prove $\lim_{x \to 2} (3x - 1) = 5$

Proof:

Given $\epsilon > 0$, we want to find $\delta > 0$ such that:

If $0 < |x - 2| < \delta$, then $|(3x - 1) - 5| < \epsilon$

Simplify: $|3x - 6| = 3|x - 2| < \epsilon$

So we need: $|x - 2| < \frac{\epsilon}{3}$

Choose $\delta = \frac{\epsilon}{3}$

Then if $0 < |x - 2| < \delta$:

$|(3x - 1) - 5| = 3|x - 2| < 3\delta = 3 \cdot \frac{\epsilon}{3} = \epsilon$

Thus, $\lim_{x \to 2} (3x - 1) = 5$

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3. Basic Limit Theorems

3.1 Limit Laws

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

1. Sum/Difference: $\lim_{x \to a} [f(x) \pm g(x)] = L \pm M$

2. Constant Multiple: $\lim_{x \to a} [c \cdot f(x)] = c \cdot L$

3. Product: $\lim_{x \to a} [f(x) \cdot g(x)] = L \cdot M$

4. Quotient: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ (provided $M \neq 0$)

5. Power: $\lim_{x \to a} [f(x)]^n = L^n$ for any positive integer $n$

6. Root: $\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{L}$ (for $n$ even, require $L \geq 0$)

3.2 Important Special Limits

1. $\lim_{x \to a} c = c$ (constant function)

2. $\lim_{x \to a} x = a$

3. $\lim_{x \to a} x^n = a^n$

4. $\lim_{x \to a} \sqrt[n]{x} = \sqrt[n]{a}$ (with appropriate conditions)

3.3 Example Using Limit Laws

Find $\lim_{x \to 3} (2x^2 - 5x + 1)$

Solution:

Using limit laws:

$$\lim_{x \to 3} (2x^2 - 5x + 1) = 2\lim_{x \to 3} x^2 - 5\lim_{x \to 3} x + \lim_{x \to 3} 1$$

$$= 2(3^2) - 5(3) + 1 = 18 - 15 + 1 = 4$$

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4. Techniques for Evaluating Limits

4.1 Direct Substitution

If $f(x)$ is defined at $x = a$ and is continuous there, then:

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

Example: $\lim_{x \to 2} (x^2 + 3x - 1) = 2^2 + 3(2) - 1 = 4 + 6 - 1 = 9$

4.2 Factoring

Use when substitution gives $\frac{0}{0}$ (indeterminate form).

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

Direct substitution gives $\frac{0}{0}$ (indeterminate).

Factor: $\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$ for $x \neq 2$

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

4.3 Rationalizing

Use for expressions with radicals.

Example: Find $\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$

Direct substitution gives $\frac{0}{0}$.

Rationalize numerator:

$$\frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \frac{(x+1) - 1}{x(\sqrt{x+1} + 1)}$$

$$= \frac{x}{x(\sqrt{x+1} + 1)} = \frac{1}{\sqrt{x+1} + 1}$$

So $\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} = \frac{1}{\sqrt{0+1} + 1} = \frac{1}{2}$

4.4 Using Special Limits

Important trigonometric limits:

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

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

3. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

Example: Find $\lim_{x \to 0} \frac{\sin 3x}{x}$

Rewrite: $\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}$

Let $u = 3x$, as $x \to 0$, $u \to 0$

So $\lim_{x \to 0} \frac{\sin 3x}{x} = 3 \cdot \lim_{u \to 0} \frac{\sin u}{u} = 3 \cdot 1 = 3$

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5. Limits at Infinity

5.1 Definition

• $\lim_{x \to \infty} f(x) = L$: As $x$ increases without bound, $f(x)$ approaches $L$

• $\lim_{x \to -\infty} f(x) = L$: As $x$ decreases without bound, $f(x)$ approaches $L$

5.2 Basic Limits at Infinity

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

2. $\lim_{x \to -\infty} \frac{1}{x} = 0$

3. $\lim_{x \to \infty} \frac{1}{x^n} = 0$ for $n > 0$

4. $\lim_{x \to \infty} e^x = \infty$

5. $\lim_{x \to -\infty} e^x = 0$

5.3 Rational Functions at Infinity

For $f(x) = \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0}{b_m x^m + b_{m-1} x^{m-1} + \cdots + b_0}$:

1. If $n < m$: $\lim_{x \to \infty} f(x) = 0$

2. If $n = m$: $\lim_{x \to \infty} f(x) = \frac{a_n}{b_m}$

3. If $n > m$: $\lim_{x \to \infty} f(x) = \pm \infty$ (sign depends on leading coefficients)

Example: Find $\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{5x^2 + 4x - 3}$

Divide numerator and denominator by $x^2$:

$$\lim_{x \to \infty} \frac{3 - \frac{2}{x} + \frac{1}{x^2}}{5 + \frac{4}{x} - \frac{3}{x^2}} = \frac{3 - 0 + 0}{5 + 0 - 0} = \frac{3}{5}$$

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6. Infinite Limits

6.1 Definition

• $\lim_{x \to a} f(x) = \infty$: As $x$ approaches $a$, $f(x)$ increases without bound

• $\lim_{x \to a} f(x) = -\infty$: As $x$ approaches $a$, $f(x)$ decreases without bound

Precise definition: $\lim_{x \to a} f(x) = \infty$ means:

For every $M > 0$, there exists $\delta > 0$ such that:

If $0 < |x - a| < \delta$, then $f(x) > M$

6.2 Examples

Example 1: $\lim_{x \to 0} \frac{1}{x^2} = \infty$

As $x \to 0$, $\frac{1}{x^2}$ becomes very large positive.

Example 2: $\lim_{x \to 0^+} \frac{1}{x} = \infty$, $\lim_{x \to 0^-} \frac{1}{x} = -\infty$

Note: The two-sided limit $\lim_{x \to 0} \frac{1}{x}$ does not exist.

6.3 Vertical Asymptotes

The line $x = a$ is a vertical asymptote of $f(x)$ if at least one of these is true:

$\lim_{x \to a^+} f(x) = \pm \infty$

$\lim_{x \to a^-} f(x) = \pm \infty$

Example: $f(x) = \frac{1}{x-2}$ has vertical asymptote at $x = 2$ because:

$\lim_{x \to 2^+} \frac{1}{x-2} = \infty$ and $\lim_{x \to 2^-} \frac{1}{x-2} = -\infty$

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7. Continuity

7.1 Definition of Continuity

A function $f$ is continuous at a point $a$ if:

1. $f(a)$ is defined

2. $\lim_{x \to a} f(x)$ exists

3. $\lim_{x \to a} f(x) = f(a)$

If any condition fails, $f$ is discontinuous at $a$.

7.2 Types of Discontinuity

a) Removable Discontinuity (Hole)

$\lim_{x \to a} f(x)$ exists but is not equal to $f(a)$ (or $f(a)$ is undefined).

Example: $f(x) = \frac{x^2 - 1}{x - 1}$ has removable discontinuity at $x = 1$

Here $\lim_{x \to 1} f(x) = 2$ but $f(1)$ is undefined.

b) Jump Discontinuity

The left and right limits exist but are not equal.

Example:

$$f(x) = \begin{cases} x & \text{if } x < 1 \\ x + 1 & \text{if } x \geq 1 \end{cases}$$

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

These are different, so jump discontinuity.

c) Infinite Discontinuity

At least one one-sided limit is infinite.

Example: $f(x) = \frac{1}{x}$ at $x = 0$

7.3 Continuity on an Interval

• $f$ is continuous on an open interval $(a, b)$ if continuous at every point in $(a, b)$

• $f$ is continuous on a closed interval $[a, b]$ if:

  1. Continuous on $(a, b)$

  2. Continuous from right at $a$: $\lim_{x \to a^+} f(x) = f(a)$

  3. Continuous from left at $b$: $\lim_{x \to b^-} f(x) = f(b)$

7.4 Properties of Continuous Functions

If $f$ and $g$ are continuous at $a$, then:

1. $f \pm g$ is continuous at $a$

2. $f \cdot g$ is continuous at $a$

3. $\frac{f}{g}$ is continuous at $a$ (provided $g(a) \neq 0$)

4. $f \circ g$ (composition) is continuous at $a$

7.5 Continuous Functions

The following are continuous on their domains:

1. Polynomials

2. Rational functions

3. Root functions

4. Trigonometric functions

5. Exponential functions

6. Logarithmic functions

Example: $f(x) = \sqrt{x^2 + 1}$ is continuous everywhere because:

• $x^2 + 1$ is continuous everywhere (polynomial)

• Square root function is continuous on $[0, \infty)$

• Composition of continuous functions is continuous

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8. The Intermediate Value Theorem

8.1 Statement

If $f$ is continuous on $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c$ in $(a, b)$ such that:

$f(c) = N$

8.2 Interpretation

If a continuous function takes two values, it must take every value in between.

Graphically: The graph of a continuous function has no breaks, so it must cross any horizontal line between $f(a)$ and $f(b)$.

8.3 Application: Root Finding

Corollary: If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has at least one root in $(a, b)$.

Example: Show that $f(x) = x^3 - x - 1$ has a root between 1 and 2.

Check: $f(1) = 1 - 1 - 1 = -1 < 0$

$f(2) = 8 - 2 - 1 = 5 > 0$

Since $f$ is continuous (polynomial) and $f(1) < 0 < f(2)$, by IVT there exists $c$ in $(1, 2)$ such that $f(c) = 0$.

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9. Limits of Composite Functions

9.1 Theorem

If $\lim_{x \to a} g(x) = L$ and $f$ is continuous at $L$, then:

$\lim_{x \to a} f(g(x)) = f(L) = f\left(\lim_{x \to a} g(x)\right)$

9.2 Examples

Example 1: Find $\lim_{x \to 2} \sqrt{x^2 + 5}$

Since $\lim_{x \to 2} (x^2 + 5) = 9$ and square root function is continuous at 9:

$\lim_{x \to 2} \sqrt{x^2 + 5} = \sqrt{9} = 3$

Example 2: Find $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$

This limit does NOT exist because $\frac{1}{x} \to \pm \infty$ as $x \to 0$, and $\sin(\theta)$ oscillates between -1 and 1 as $\theta \to \infty$.

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10. The Squeeze Theorem

10.1 Statement

If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (except possibly at $a$), and:

$\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$

Then: $\lim_{x \to a} f(x) = L$

10.2 Applications

Example 1: Find $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

We know: $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$ for all $x \neq 0$

Multiply by $x^2$ (positive): $-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$

Now $\lim_{x \to 0} (-x^2) = 0$ and $\lim_{x \to 0} x^2 = 0$

By Squeeze Theorem: $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$

Example 2: Prove $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Geometric proof using squeeze theorem and unit circle.

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11. L'Hôpital's Rule

11.1 Statement

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, and $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists, then:

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

Important: Only applies to $\frac{0}{0}$ or $\frac{\infty}{\infty}$ forms.

11.2 Examples

Example 1: Find $\lim_{x \to 0} \frac{\sin x}{x}$ ($\frac{0}{0}$ form)

Apply L'Hôpital's Rule:

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

Example 2: Find $\lim_{x \to \infty} \frac{x}{e^x}$ ($\frac{\infty}{\infty}$ form)

Apply L'Hôpital's Rule:

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

Example 3: Find $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ ($\frac{0}{0}$ form)

Apply L'Hôpital's Rule twice:

First application: $\lim_{x \to 0} \frac{\sin x}{2x}$

Second application: $\lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}$

11.3 Other Indeterminate Forms

Other forms can be converted to $\frac{0}{0}$ or $\frac{\infty}{\infty}$:

1. $0 \cdot \infty$: Rewrite as $\frac{0}{1/\infty}$ or $\frac{\infty}{1/0}$

2. $\infty - \infty$: Combine into single fraction

3. $0^0$, $\infty^0$, $1^\infty$: Use logarithm: $\lim f(x)^{g(x)} = e^{\lim g(x) \ln f(x)}$

Example: $\lim_{x \to 0^+} x^x$ ($0^0$ form)

Let $y = x^x$, then $\ln y = x \ln x$

$\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}$ ($\frac{\infty}{\infty}$ form)

Apply L'Hôpital's Rule: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0$

So $\lim_{x \to 0^+} \ln y = 0$, thus $\lim_{x \to 0^+} y = e^0 = 1$

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12. Practical Applications

12.1 Instantaneous Rate of Change

The derivative: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

Example: Find instantaneous velocity at $t = 2$ if position $s(t) = t^2$

$s'(2) = \lim_{h \to 0} \frac{(2+h)^2 - 2^2}{h} = \lim_{h \to 0} \frac{4 + 4h + h^2 - 4}{h}$

$= \lim_{h \to 0} \frac{4h + h^2}{h} = \lim_{h \to 0} (4 + h) = 4$

12.2 Area Under a Curve

Definite integral as limit of Riemann sums:

$\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$

12.3 Continuity in Real World

• Temperature change over time

• Flow of water in a pipe

• Motion of objects (position as continuous function of time)

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13. Important Limits to Memorize

13.1 Basic Limits

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

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

3. $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$

4. $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

5. $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$

6. $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$

7. $\lim_{x \to 0} (1 + x)^{1/x} = e$

13.2 Trigonometric Limits

1. $\lim_{x \to 0} \frac{\tan x}{x} = 1$

2. $\lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1$

3. $\lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1$

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14. Solved Examples

Example 1: Piecewise Function Limit

Find $\lim_{x \to 2} f(x)$ where:

$$f(x) = \begin{cases} x^2 & \text{if } x < 2 \\ x + 2 & \text{if } x \geq 2 \end{cases}$$

Solution: Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x^2 = 4$

Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x+2) = 4$

Since both equal 4, $\lim_{x \to 2} f(x) = 4$

Example 2: Rational Function at Infinity

Find $\lim_{x \to \infty} \frac{2x^3 - x + 1}{5x^3 + 3x^2 - 2}$

Solution: Divide numerator and denominator by $x^3$:

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

Example 3: Using Conjugates

Find $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$

Solution: Multiply numerator and denominator by conjugate:

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

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

Example 4: Squeeze Theorem Application

Find $\lim_{x \to \infty} \frac{\cos x}{x}$

Solution: We know: $-1 \leq \cos x \leq 1$ for all $x$

Divide by $x > 0$: $-\frac{1}{x} \leq \frac{\cos x}{x} \leq \frac{1}{x}$

Now $\lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$

By Squeeze Theorem: $\lim_{x \to \infty} \frac{\cos x}{x} = 0$

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15. Common Mistakes and Exam Tips

15.1 Common Mistakes

1. Assuming limit exists when one-sided limits differ

   • Always check both sides for piecewise functions and functions with absolute values

2. Applying L'Hôpital's Rule to non-indeterminate forms

   • Check it's $\frac{0}{0}$ or $\frac{\infty}{\infty}$ first

3. Confusing $\lim_{x \to a} f(x)$ with $f(a)$

   • They're equal only if $f$ is continuous at $a$

4. Incorrect handling of infinity in limits

   • Remember: $\infty + \infty = \infty$, but $\infty - \infty$ is indeterminate

5. Forgetting to check domain restrictions

   • Especially for rational functions (denominator ≠ 0) and even roots (radicand ≥ 0)

15.2 Problem-Solving Strategy

1. Try direct substitution first

2. If indeterminate (0/0, ∞/∞, etc.):

   • Factor and cancel

   • Rationalize

   • Use special limits

   • Apply L'Hôpital's Rule (if appropriate)

3. For piecewise functions: Check left and right limits separately

4. For limits at infinity: Divide by highest power

5. Always verify your answer makes sense

15.3 Quick Checks

1. Continuity check: $\lim_{x \to a} f(x) = f(a)$?

2. One-sided limits: Equal for two-sided limit to exist

3. Indeterminate forms: Recognize and handle properly

4. Special limits: Know the important ones

5. Squeeze Theorem: Useful for oscillating functions bounded by known functions

This comprehensive theory covers all aspects of limits and continuity with detailed explanations and examples, providing complete preparation for the entrance examination.