Limits, Continuity, and Their Applications

1. Introduction to Limits

1.1 Definition of Limit

The limit of a function $f(x)$ as $x$ approaches a value $a$ is the value that $f(x)$ gets arbitrarily close to as $x$ approaches $a$. Mathematically, we write:

$$\underset{x\to a}{\lim }f(x)=L$$

if for every $ϵ>0$, there exists a $\delta >0$ such that:

$$0<|x-a|<\delta ⟹|f(x)-L|<ϵ$$

Key Points:

• The limit $L$ must exist independently of how $x$ approaches $a$ (from left or right).
• The function $f(x)$ need not be defined at $x=a$ for the limit to exist.

1.2 Types of Limits

• Finite Limits: ${\lim }_{x\to a}f(x)=L$ where $L$ is finite.
• Infinite Limits: ${\lim }_{x\to a}f(x)=\infty$ or $-\infty$.
• One-Sided Limits:
  • Left-hand limit: ${\lim }_{x\to a^{-}}f(x)$
  • Right-hand limit: ${\lim }_{x\to a^{+}}f(x)$
• Limits at Infinity: ${\lim }_{x\to \infty }f(x)$ or ${\lim }_{x\to -\infty }f(x)$.

1.3 Evaluating Limits

Direct Substitution Method

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

$$\underset{x\to a}{\lim }f(x)=f(a)$$

Example: Evaluate ${\lim }_{x\to 2}(3x^2-4x+5)$.

Solution: Since the function is a polynomial (continuous everywhere), we substitute $x=2$:

$$3(2{)}^2-4(2)+5=12-8+5=9$$

Factoring Method

If direct substitution results in $\frac{0}{0}$, factor the numerator and denominator.

Example: Evaluate ${\lim }_{x\to 3}\frac{x^2-9}{x-3}$.

Solution: Factor the numerator:

$$\frac{(x-3)(x+3)}{x-3}=x+3\text{(for x=3)}$$

Now, substitute $x=3$:

$$3+3=6$$

Rationalizing Method

For limits involving square roots, multiply by the conjugate.

Example: Evaluate ${\lim }_{x\to 0}\frac{\sqrt{x+4}-2}{x}$.

Solution: Multiply numerator and denominator by the conjugate $\sqrt{x+4}+2$:

$$\frac{(\sqrt{x+4}-2)(\sqrt{x+4}+2)}{x(\sqrt{x+4}+2)}=\frac{(x+4)-4}{x(\sqrt{x+4}+2)}=\frac{x}{x(\sqrt{x+4}+2)}=\frac{1}{\sqrt{x+4}+2}$$

Now, substitute $x=0$:

$$\frac{1}{2+2}=\frac{1}{4}$$

Limits at Infinity

For rational functions $\frac{P(x)}{Q(x)}$, compare the degrees of $P(x)$ and $Q(x)$:

• If $\mathrm{deg}(P)<\mathrm{deg}(Q)$, ${\lim }_{x\to \infty }\frac{P(x)}{Q(x)}=0$.
• If $\mathrm{deg}(P)=\mathrm{deg}(Q)$, ${\lim }_{x\to \infty }\frac{P(x)}{Q(x)}=\frac{\text{leading coefficient of }P}{\text{leading coefficient of }Q}$.
• If $\mathrm{deg}(P)>\mathrm{deg}(Q)$, the limit is $\pm \infty$.

Example: Evaluate ${\lim }_{x\to \infty }\frac{3x^2+2x-1}{2x^2-5}$.

Solution: Divide numerator and denominator by $x^2$:

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

1.4 Special Limits

1. Limit of $\frac{\sin x}{x}$ as $x\to 0$: $$\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$$

2. Limit of $\frac{1-\cos x}{x}$ as $x\to 0$: $$\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$$

3. Limit of $\frac{e^x-1}{x}$ as $x\to 0$: $$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$

Example: Evaluate ${\lim }_{x\to 0}\frac{\sin 5x}{3x}$.

Solution: Use the special limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$:

$$\underset{x\to 0}{\lim }\frac{\sin 5x}{3x}=\frac{5}{3}\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}=\frac{5}{3}\times 1=\frac{5}{3}$$

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2. Continuity of Functions

2.1 Definition of Continuity at a Point

A function $f(x)$ is continuous at $x=a$ if the following three conditions are met:

1. $f(a)$ is defined.
2. ${\lim }_{x\to a}f(x)$ exists.
3. ${\lim }_{x\to a}f(x)=f(a)$.

If any of these conditions fail, $f(x)$ is discontinuous at $x=a$.

2.2 Types of Discontinuities

2.3 Continuity on an Interval

A function $f(x)$ is continuous on an interval if it is continuous at every point in that interval. For closed intervals $[a,b]$:

• $f(x)$ must be continuous on $(a,b)$.
• ${\lim }_{x\to a^{+}}f(x)=f(a)$ (right-hand continuity at $a$).
• ${\lim }_{x\to b^{-}}f(x)=f(b)$ (left-hand continuity at $b$).

2.4 Testing Continuity

To check if $f(x)=\sqrt{1-x^2}$ is continuous on $[-1,1]$:

1. Domain Check: The expression under the square root must be non-negative: $$1-x^2\ge 0⟹x^2\le 1⟹-1\le x\le 1$$ So, $f(x)$ is defined on $[-1,1]$.

2. Continuity at Endpoints:

   • At $x=-1$: $$f(-1)=\sqrt{1-(-1{)}^2}=0$$ $$\underset{x\to -1^{+}}{\lim }f(x)=0=f(-1)$$
   • At $x=1$: $$f(1)=\sqrt{1-1^2}=0$$ $$\underset{x\to 1^{-}}{\lim }f(x)=0=f(1)$$

3. Continuity on $(-1,1)$: The function is a composition of continuous functions (square root and polynomial), so it is continuous on $(-1,1)$.

Conclusion: $f(x)$ is continuous on $[-1,1]$.

2.5 Intermediate Value Theorem (IVT)

If $f(x)$ is continuous on $[a,b]$ and $N$ is any number between $f(a)$ and $f(b)$, then there exists a $c\in (a,b)$ such that $f(c)=N$.

Example: Show that $f(x)=x^3-x-2$ has a root in $[1,2]$.

Solution:

• $f(1)=1-1-2=-2$
• $f(2)=8-2-2=4$ Since $f(x)$ is continuous on $[1,2]$ and $0$ is between $-2$ and $4$, by IVT, there exists $c\in (1,2)$ such that $f(c)=0$.

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3. Applications of Continuity

3.1 Evaluating Limits Using Continuity

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

$$\underset{x\to a}{\lim }f(x)=f(a)$$

Example: Evaluate ${\lim }_{x\to 4}\frac{5+\sqrt{x}}{\sqrt{5+x}}$.

Solution: Let $f(x)=\frac{5+\sqrt{x}}{\sqrt{5+x}}$. Check continuity at $x=4$:

• $f(4)=\frac{5+2}{\sqrt{9}}=\frac{7}{3}$.
• The denominator $\sqrt{5+x}$ is defined and non-zero near $x=4$, so $f(x)$ is continuous at $x=4$.

Thus:

$$\underset{x\to 4}{\lim }f(x)=f(4)=\frac{7}{3}$$

3.2 Removing Discontinuities

If a function has a removable discontinuity at $x=a$, we can redefine $f(a)$ to make it continuous.

Example: Let $f(x)=\frac{x^2-1}{x-1}$. The function is undefined at $x=1$, but the limit exists:

$$\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}=\underset{x\to 1}{\lim }(x+1)=2$$

Redefine $f(1)=2$ to make $f(x)$ continuous at $x=1$.

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4. Common Mistakes and Pitfalls

1. Assuming Continuity Everywhere: Not all functions are continuous (e.g., $f(x)=\frac{1}{x}$ at $x=0$).
2. Ignoring Domain Restrictions: For $\sqrt{x}$, $x\ge 0$. For $\frac{1}{x}$, $x\ne 0$.
3. Incorrect Limit Evaluation: Always check if direct substitution is valid. If not, use algebraic manipulation.
4. Misapplying IVT: The function must be continuous on the entire interval $[a,b]$.

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5. Exam Tips

5.1 For Limit Questions

• Always check the domain of the function before evaluating limits.
• Factor, rationalize, or use special limits when direct substitution fails.
• For limits at infinity, compare the highest degree terms in the numerator and denominator.
• Practice one-sided limits (e.g., $x\to a^{+}$ or $x\to a^{-}$) separately.

5.2 For Continuity Questions

• Verify all three conditions for continuity at a point: $f(a)$ defined, limit exists, and limit equals $f(a)$.
• For interval continuity, check endpoints separately (left-hand and right-hand limits).
• Use IVT carefully: Ensure the function is continuous on the entire closed interval.
• Identify types of discontinuities (removable, jump, infinite, essential) and explain why they occur.

5.3 Common Exam Patterns

1. Define continuity at a point and verify it for a given function.
2. Evaluate limits using algebraic techniques or continuity.
3. Determine continuity on an interval and identify discontinuities.
4. Apply IVT to prove the existence of roots or intermediate values.

5.4 Model Answer Structure

• Definitions: Clearly state definitions with mathematical notation.
• Examples: Show step-by-step solutions with explanations.
• Graphical Interpretation: Sketch graphs where helpful (e.g., jump discontinuities).
• Conclusion: Summarize findings concisely.

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