Limits And Continuity (SAQs)

Maths-1B | 8. Limits And Continuity – SAQs:
Welcome to SAQs in Chapter 8: Limits And Continuity. This page contains the most Important FAQs for Short Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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SAQ-1 : Is f defined by f(x) = {sin2x/x if x ≠0 if x = 0 continuous at 0?

$$\text{Given } f(0) = 1 \quad \text{(1)}$$

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{(2)}$$

$$\text{From (1) \& (2) } \lim_{x \to 0} f(x) \neq f(0)$$

$$\text{Hence proved that } f \text{ is not continuous at } x = 0$$

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SAQ-2 : Is f given by f(x) = {x² – 9/x² – 2x – 3 if 0<x<5, x≠0 if x = 3

$$\text{Given } f(3) = 1.5 \quad \text{(1)}$$

$$\lim_{x \to 3} \frac{x^2 – 9}{x^2 – 2x – 3} = \lim_{x \to 3} \frac{(x – 3)(x + 3)}{(x – 3)(x + 1)} = \lim_{x \to 3} \frac{x + 3}{x + 1}$$

$$\lim_{x \to 3} \frac{x + 3}{x + 1} = \frac{3 + 3}{3 + 1} = \frac{6}{4} = \frac{3}{2} = 1.5 \quad \text{(2)}$$

$$\text{From (1) \& (2) } \lim_{x \to 3} f(x) = f(3)$$

$$\text{Hence proved that } f(x) \text{ is continuous at } x = 3$$

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SAQ-3 : If f is given by f(x) = {k²x – k 2 if x ≥ 1 if x < 1 is a continuous on R, then find k

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

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (k^2x – k) = k^2 \cdot 1 – k = k^2 – k \quad \text{(2)}$$

$$\text{From (1) \& (2), L.H.L = R.H.L}$$

$$k^2 – k = 2$$

$$k^2 – k – 2 = 0$$

$$k^2 – k – 2 = 0$$

$$(k – 2)(k + 1) = 0$$

$$k = 2 \quad \text{or} \quad k = -1$$

$$\text{Hence, } k = 2 \quad \text{or} \quad k = -1$$

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SAQ-4 : Check the continuity of f(x) = {1/2(x² – 4) if 0 < x < 2 {0 if x = 2 {2 – 8x^-3 if x > 2 at 2

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1}{2}(x^2 – 4) = \frac{1}{2}(4 – 4) = 0 \quad \text{(1)}$$

$$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2 – \frac{8}{x^3}) = 2 – \frac{8}{8} = 2 – 1 = 1 \quad \text{(2)}$$

$$\text{From (1) \& (2), L.H.L} \neq \text{R.H.L}$$

$$\text{Hence proved that } f(x) \text{ is not continuous at } x = 2$$

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SAQ-5 : Show that f(x) = {cos ax – cos bx/x² if x ≠ 0 1/2(b² – a²) if x = 0 is continuous at 0

$$\text{Given } f(0) = \frac{b^2 – a^2}{2} \quad \text{(1)}$$

$$\lim_{x \to 0} \frac{\cos(ax) – \cos(bx)}{x^2}$$

$$\lim_{x \to 0} \frac{-2 \sin\left(\frac{ax + bx}{2}\right) \sin\left(\frac{ax – bx}{2}\right)}{x^2}$$

$$= -2 \lim_{x \to 0} \left(\frac{\sin\left(\frac{a + b}{2}x\right)}{x}\right)\left(\frac{\sin\left(\frac{a – b}{2}x\right)}{x}\right)$$

$$= -2 \left(\lim_{x \to 0} \frac{\sin\left(\frac{a + b}{2}x\right)}{x}\right)\left(\lim_{x \to 0} \frac{\sin\left(\frac{a – b}{2}x\right)}{x}\right)$$

$$= -2 \left(\frac{a + b}{2}\right)\left(\frac{a – b}{2}\right) = -\left(\frac{a^2 – b^2}{2}\right)$$

$$= \frac{b^2 – a^2}{2} \quad \text{(2)}$$

$$\text{From (1) \& (2) }$$

$$\text{Hence proved that } f(x) \text{ is continuous at } x = 0$$

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SAQ-6 : Find the real constants a, b so that the function f given by f(x) = {sin x if x ≤ 0 x² + a if 0 < x < 1 bx + 3 if 1 ≤ x ≤ 3 -3 if x > 3 is continuous on R

Given that f(x) is continuous on R ⇒ f(x) is continuous at x = 0, 3

Part (i): Continuity at x = 0

Left-hand limit (L.H.L) as 𝑥 approaches 0 from the left:

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

Right-hand limit (R.H.L) as x approaches 0 from the right:

$$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + a) = 0^2 + a = a$$

Equate L.H.L and R.H.L for continuity:

$$0 = a$$

$$\Rightarrow a = 0$$

Part (ii): Continuity at x = 3

Left-hand limit (L.H.L) as x approaches 3 from the left:

$$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (bx + 3) = 3b + 3$$

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

Equate L.H.L and R.H.L for continuity:

$$3b + 3 = -3$$

$$3b = -6 \Rightarrow b = -2$$

$$\Rightarrow b = -2$$

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SAQ-7 : Evaluate Lt x→0 cos ax – cos bx/x²

$$\lim_{x \to 0} \frac{\cos ax – \cos bx}{x^2} = \lim_{x \to 0} \frac{2 \sin\left(\frac{ax + bx}{2}\right) \sin\left(\frac{bx – ax}{2}\right)}{x^2}$$

$$= 2 \lim_{x \to 0} \frac{\sin\left(\frac{a + b}{2}x\right)}{x} \cdot \frac{\sin\left(\frac{b – a}{2}x\right)}{x}$$

$$= 2 \left(\frac{a + b}{2}\right) \left(\frac{b – a}{2}\right)$$

$$= 2 \left(\frac{a + b}{2}\right) \left(\frac{b – a}{2}\right) = \frac{(a+b)(b-a)}{2} = \frac{ab – a^2 + ba – b^2}{2} = \frac{b^2 – a^2}{2}$$

$$\lim_{x \to 0} \frac{\cos ax – \cos bx}{x^2} = \frac{b^2 – a^2}{2}$$

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SAQ-8 : Find Lt x→0 (x sin a – a sin x/x – a)

$$\lim_{x \to a} \frac{x \sin a – a \sin x}{x – a}$$

$$\lim_{x \to a} \frac{x \sin a – a \sin a + a \sin a – a \sin x}{x – a}$$

$$= \lim_{x \to a} \sin a \frac{(x – a)}{x – a} – a \lim_{x \to a} \frac{\sin x – \sin a}{x – a}$$

$$\sin x – \sin a = 2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x – a}{2}\right)$$

$$-a \lim_{x \to a} \frac{2 \cos\left(\frac{x + a}{2}\right) \sin\left(\frac{x – a}{2}\right)}{x – a}$$

$$-a \cdot 2 \cos\left(a + \frac{a}{2}\right) \cdot \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$$

$$= \sin a – 2a \cos\left(\frac{3a}{2}\right) \cdot \frac{1}{2}$$

$$= \sin a – a \cos\left(\frac{3a}{2}\right)$$

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SAQ-9 : Evaluate Lt x→0 e^x – sinx – 1/x

$$\lim_{x \to 0} \frac{e^x – \sin x – 1}{x}$$

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

$$\lim_{x \to 0} \frac{e^x – 1}{x} = \lim_{x \to 0} \frac{1 + x + \frac{x^2}{2} + O(x^3) – 1}{x} = \lim_{x \to 0} \left(1 + \frac{x}{2} + O(x^2)\right) = 1$$

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

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

$$= 1 – 1 = 0$$