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Calculus S.M Yusuf Chapter 1 Ex 1.1 Solved Notes & Quiz
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Calculus S.M Yusuf Chapter 1 Ex 1.1 Solved Notes & Quiz
In calculus and real analysis, solving inequalities and evaluating limits form the foundation for understanding continuity, derivatives, and integrals. The key definitions and theorems governing Exercise 1.1 are outlined below:
| Concept / Identity | Mathematical Formula | Key Application |
|---|---|---|
| Absolute Value Inequality (Bounded) | $|ax + b| \le c \iff -c \le ax + b \le c$ | Finding domain / interval solutions |
| Absolute Value Inequality (Unbounded) | $|ax + b| \ge c \iff ax + b \ge c \text{ or } ax + b \le -c$ | Union of disjoint intervals |
| Standard Squeeze Limit | $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$ | Evaluating trigonometric limits |
| Cosine Limit Identity | $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ | Indeterminate forms of $0/0$ |
| Rational Limit at Infinity | $\lim_{x \to \infty} \frac{a_n x^n + \dots}{b_n x^n + \dots} = \frac{a_n}{b_n}$ | Horizontal asymptotes / Limits at $\infty$ |
Question 1 (Part i): Solve the absolute value inequality $|2x - 5| < 7$ and express the solution set in interval notation.
Solution:
Using the absolute value property $|u| < a \iff -a < u < a$ for $a > 0$:
$$-7 < 2x - 5 < 7$$
Add $5$ to all three parts of the inequality:
$$-7 + 5 < 2x < 7 + 5$$
$$-2 < 2x < 12$$
Divide all parts by $2$:
$$-1 < x < 6$$
In interval notation, the solution set is $(-1, 6)$.
Answer: $$(-1, 6)$$
Question 1 (Part ii): Solve the absolute value inequality $|3x + 4| \ge 10$ and write the solution set in interval notation.
Solution:
Using the absolute value property $|u| \ge a \iff u \ge a \text{ or } u \le -a$:
$$\text{Case 1: } 3x + 4 \ge 10 \implies 3x \ge 6 \implies x \ge 2$$
$$\text{Case 2: } 3x + 4 \le -10 \implies 3x \le -14 \implies x \le -\frac{14}{3}$$
Combining both cases gives $x \in \left(-\infty, -\frac{14}{3}\right] \cup [2, \infty)$.
Answer: $$\left(-\infty, -\frac{14}{3}\right] \cup [2, \infty)$$
Question 2 (Part i): Solve the quadratic inequality $x^2 - 3x - 10 \le 0$.
Solution:
First, factor the quadratic expression:
$$x^2 - 3x - 10 = (x - 5)(x + 2) \le 0$$
Find the critical numbers by setting $(x - 5)(x + 2) = 0$, which gives $x = -2$ and $x = 5$.
These points divide the real number line into three test intervals: $(-\infty, -2)$, $(-2, 5)$, and $(5, \infty)$.
Question 2 (Part ii): Solve the rational inequality $\frac{x + 2}{x - 4} > 0$.
Solution:
Find the boundary points where the numerator is zero or the denominator is zero:
$$\text{Numerator } x + 2 = 0 \implies x = -2$$
$$\text{Denominator } x - 4 = 0 \implies x = 4$$
The critical numbers are $x = -2$ and $x = 4$. Test the three intervals $(-\infty, -2)$, $(-2, 4)$, and $(4, \infty)$:
Question 3 (Part i): Evaluate the limit $\lim_{x \to 3} \frac{x^2 - 9}{x - 3}$.
Solution:
Direct substitution of $x = 3$ yields the indeterminate form $\frac{3^2 - 9}{3 - 3} = \frac{0}{0}$.
Factor the numerator using the difference of squares $a^2 - b^2 = (a - b)(a + b)$:
$$\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3}$$
Since $x \to 3$, $x \neq 3$, so we can cancel the non-zero common factor $(x - 3)$:
$$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$$
Answer: $$6$$
Question 3 (Part ii): Evaluate the limit $\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$.
Solution:
Direct substitution gives $\frac{2^3 - 8}{2^2 - 4} = \frac{0}{0}$.
Factor numerator using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and denominator using $a^2 - b^2 = (a - b)(a + b)$:
$$\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)}$$
Cancel the common term $(x - 2)$:
$$\lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 2} = \frac{(2)^2 + 2(2) + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3$$
Answer: $$3$$
Question 4 (Part i): Evaluate the trigonometric limit $\lim_{x \to 0} \frac{\sin(5x)}{3x}$.
Solution:
Direct substitution yields $\frac{0}{0}$. Use the fundamental limit theorem $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$.
Multiply and divide by $5$:
$$\lim_{x \to 0} \frac{\sin(5x)}{3x} = \lim_{x \to 0} \left( \frac{5}{3} \cdot \frac{\sin(5x)}{5x} \right)$$
Let $\theta = 5x$. As $x \to 0$, $\theta \to 0$:
$$= \frac{5}{3} \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = \frac{5}{3} (1) = \frac{5}{3}$$
Answer: $$\frac{5}{3}$$
Question 4 (Part ii): Evaluate the limit $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.
Solution:
Using the trigonometric double-angle identity $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$:
$$\lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2}$$
Rewrite the expression:
$$= 2 \lim_{x \to 0} \left[ \frac{\sin(x/2)}{x} \right]^2 = 2 \lim_{x \to 0} \left[ \frac{1}{2} \cdot \frac{\sin(x/2)}{x/2} \right]^2$$
Since $\lim_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$:
$$= 2 \left( \frac{1}{2} \cdot 1 \right)^2 = 2 \cdot \frac{1}{4} = \frac{1}{2}$$
Answer: $$\frac{1}{2}$$
Question 5 (Part i): Evaluate the limit at infinity $\lim_{x \to \infty} \frac{4x^2 - 3x + 7}{2x^2 + 5}$.
Solution:
Divide both numerator and denominator by the highest power of $x$ present in the denominator, which is $x^2$:
$$\lim_{x \to \infty} \frac{\frac{4x^2}{x^2} - \frac{3x}{x^2} + \frac{7}{x^2}}{\frac{2x^2}{x^2} + \frac{5}{x^2}} = \lim_{x \to \infty} \frac{4 - \frac{3}{x} + \frac{7}{x^2}}{2 + \frac{5}{x^2}}$$
Apply the limit property $\lim_{x \to \infty} \frac{1}{x^n} = 0$ for $n > 0$:
$$= \frac{4 - 0 + 0}{2 + 0} = \frac{4}{2} = 2$$
Answer: $$2$$
Question 5 (Part ii): Evaluate the limit $\lim_{x \to \infty} \left(\sqrt{x^2 + 4x} - x\right)$.
Solution:
This has the indeterminate form $\infty - \infty$. Rationalize the expression by multiplying and dividing by its conjugate $(\sqrt{x^2 + 4x} + x)$:
$$\lim_{x \to \infty} \frac{(\sqrt{x^2 + 4x} - x)(\sqrt{x^2 + 4x} + x)}{\sqrt{x^2 + 4x} + x}$$