Limits And Continuity (VSAQs)

Maths-1B | 8. Limits And Continuity – VSAQs:
Welcome to VSAQs in Chapter 8: Limits And Continuity. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Compute Lt x→a x² – a²/x – a

$$Lt x→a \frac{x^2 – a^2}{x – a}$$

$$= Lt x→a \frac{(x + a)(x – a)}{x – a}$$

$$= Lt x→a (x + a)$$

$$= a + a = 2a$$

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VSAQ-2 : Evaluate Lt x→2 x – 2/x³ – 8

$$Lt x→2 \frac{x – 2}{x^3 – 8}$$

$$= Lt x→2 \frac{x – 2}{(x – 2)(x^2 + 2x + 4)}$$

$$= Lt x→2 \frac{1}{x^2 + 2x + 4}$$

$$= \frac{1}{2^2 + 2(2) + 4}$$

$$= \frac{1}{4 + 4 + 4} = \frac{1}{12}$$

$$Lt x→2 \frac{x – 2}{x^3 – 8} = \frac{1}{12}$$

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VSAQ-3 : Evaluate Lt x→3 x² – 8x + 15/x² – 9

$$Lt x→3 \frac{x^2 – 8x + 15}{x^2 – 9}$$

$$= Lt x→3 \frac{(x – 5)(x – 3)}{(x – 3)(x + 3)}$$

$$= Lt x→3 \frac{x – 5}{x + 3}$$

$$= \frac{3 – 5}{3 + 3}$$

$$= \frac{-2}{6} = -\frac{1}{3}$$

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VSAQ-4 : Evaluate Lt x→3 x³ – 6x² + 9x/x² – 9

$$Lt x→3 \frac{x^3 – 6x^2 + 9x}{x^2 – 9}$$

$$= Lt x→3 \frac{x(x – 3)^2}{(x – 3)(x + 3)}$$

$$= Lt x→3 \frac{x(x – 3)}{x + 3}$$

$$= \frac{3(3 – 3)}{3 + 3}$$

$$= \frac{3(0)}{6} = 0$$

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VSAQ-5 : Evaluate Lt x→3 x³ – 3x²/x² – 5x + 6

$$Lt x→3 \frac{x^3 – 3x^2}{x^2 – 5x + 6}$$

$$= Lt x→3 \frac{x^2(x – 3)}{(x – 3)(x – 2)}$$

$$= Lt x→3 \frac{x^2}{x – 2}$$

$$= \frac{3^2}{3 – 2} = \frac{9}{1} = 9$$

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VSAQ-6 : Evaluate Lt x→2 [1/x – 2 – 4/x² – 4]

$$Lt x→2 \left[\frac{1}{x – 2} – \frac{4}{x^2 – 4}\right]$$

$$= Lt x→2 \frac{(x + 2) – 4}{(x – 2)(x + 2)}$$

$$= Lt x→2 \frac{x – 2}{(x – 2)(x + 2)}$$

$$= Lt x→2 \frac{1}{x + 2}$$

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

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VSAQ-7 : Evaluate Lt x→0 √1 + x – 1/x

$$Lt x→0 \frac{\sqrt{1 + x} – 1}{x}$$

$$= Lt x→0 \frac{(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1)}{x(\sqrt{1 + x} + 1)}$$

$$= Lt x→0 \frac{x}{x(\sqrt{1 + x} + 1)}$$

$$= Lt x→0 \frac{1}{\sqrt{1 + x} + 1}$$

$$= \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$$

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VSAQ-8 : Evaluate Lt x→0 e^x – 1/√1 + x – 1

$$Lt_{x \to 0} \left( e^x – \frac{1}{\sqrt{1 + x}} – 1 \right)$$

$$= Lt_{x \to 0} \left( \frac{e^x – 1}{x} \right) \left( \frac{x}{\sqrt{1 + x} – 1} \right)$$

$$= \left( Lt_{x \to 0} \frac{e^x – 1}{x} \right) \left( Lt_{x \to 0} \frac{x}{\sqrt{1 + x} – 1} \right) = 1 \cdot Lt_{x \to 0} \frac{x}{\sqrt{1 + x} – 1}$$

$$= Lt_{x \to 0} \frac{x (\sqrt{1 + x} + 1)}{(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1)}$$

$$= Lt_{x \to 0} \frac{x (\sqrt{1 + x} + 1)}{1 + x – 1}$$

$$= Lt_{x \to 0} \frac{x (\sqrt{1 + x} + 1)}{x}$$

$$= Lt_{x \to 0} (\sqrt{1 + x} + 1)$$

$$= \sqrt{1 + 0} + 1 = 1 + 1 = 2$$

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VSAQ-9 : Find Lt x→0 3^x – 1/√1 + x – 1

$$Lt_{x \to 0} \left( 3x – \frac{1}{\sqrt{1 + x}} – 1 \right)$$

$$= Lt_{x \to 0} \left( \frac{3x – 1}{x} \right) \left( \frac{x}{\sqrt{1 + x} – 1} \right)$$

$$= \left( Lt_{x \to 0} \frac{3x – 1}{x} \right) \left( Lt_{x \to 0} \frac{x}{\sqrt{1 + x} – 1} \right) = \log 3 \cdot Lt_{x \to 0} \frac{x}{\sqrt{1 + x} – 1}$$

$$= \log 3 \cdot Lt_{x \to 0} \frac{x (\sqrt{1 + x} – 1)}{(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1)}$$

$$= \log 3 \cdot Lt_{x \to 0} \frac{x (\sqrt{1 + x} + 1)}{1 + x – 1}$$

$$= \log 3 \cdot \frac{x (\sqrt{1 + x} + 1)}{x}$$

$$= \log 3 \cdot Lt_{x \to 0} (\sqrt{1 + x} – 1)$$

$$= \log 3 \cdot (\sqrt{1 + 0} + 1) = \log 3 \cdot (1 + 1) = 2 \log 3$$

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VSAQ-10 : Evaluate Lt x→3 x³ – 27/x – 3

$$Lt_{x \to 3} \left( \frac{x^3 – 27}{x – 3} \right)$$

$$= 3 \cdot (3)^3 – 1$$

$$= 3 \cdot (3^2)$$

$$= 27$$

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VSAQ-11 : Evaluate Lt x→0 (1 + x)^3/2 – 1/x

$$Lt_{x \to 0} \left( \frac{(1 + x)^{\frac{3}{2}} – 1}{x} \right)$$

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

$$= \frac{3}{2} \cdot (1)^{\frac{3}{2}} – 1$$

$$= \frac{3}{2}$$

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VSAQ-12 : Compute Lt x→0 (sin ax/sin bx), b ≠ 0, a ≠ b

$$Lt_{x \to 0} \frac{\sin(ax)}{\sin(bx)}$$

$$= Lt_{x \to 0} \left( \frac{\sin(ax)}{x} \right) \div \left( \frac{\sin(bx)}{x} \right)$$

$$= \left( Lt_{x \to 0} \frac{\sin(ax)}{x} \right) \div \left( Lt_{x \to 0} \frac{\sin(bx)}{x} \right)$$

$$= \frac{a}{b}$$

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VSAQ-13 : Compute Lt x→0 sin ax/x cos x

$$Lt_{x \to 0} \left( \frac{\sin(ax)}{x} \cdot \cos(x) \right)$$

$$= Lt_{x \to 0} \left( \frac{\sin(ax)}{x} \right) \cdot \left( \frac{1}{\cos(x)} \right)$$

$$= a \cdot \frac{1}{\cos(0)}$$

$$= a \cdot \frac{1}{1}$$

$$= a$$

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VSAQ-14 : Evaluate Lt x→a tan(x – a)/x² – a²

$$Lt_{x \to a} \frac{\tan(x – a)}{x^2 – a^2}$$

$$= Lt_{x \to a} \frac{\tan(x – a)}{(x – a)(x + a)}$$

$$= Lt_{x \to a} \left( \frac{\tan(x – a)}{x – a} \right) \cdot \left( \frac{1}{x + a} \right)$$

$$= \left( Lt_{(x – a) \to 0} \frac{\tan(x – a)}{x – a} \right) \cdot \left( Lt_{x \to a} \frac{1}{x + a} \right)$$

$$= 1 \cdot \frac{1}{a + a}$$

$$= \frac{1}{2a}$$

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VSAQ-15 : Evaluate Lt x→1 sin(x – 1)/x² – 1

$$Lt_{x \to 1} \frac{\sin(x – 1)}{x^2 – 1}$$

$$= Lt_{x \to 1} \frac{\sin(x – 1)}{(x – 1)(x + 1)}$$

$$= Lt_{x \to 1} \left( \frac{\sin(x – 1)}{x – 1} \right) \cdot \left( \frac{1}{x + 1} \right)$$

$$= \left( Lt_{(x – 1) \to 0} \frac{\sin(x – 1)}{x – 1} \right) \cdot \left( Lt_{x \to 1} \frac{1}{x + 1} \right)$$

$$= 1 \cdot \frac{1}{1 + 1}$$

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

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VSAQ-16 : Evaluate Lt x→0 1 – cosx/x

$$Lt_{x \to 0} \frac{1 – \cos x}{x}$$

$$= Lt_{x \to 0} \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x}$$

$$= 2 \cdot Lt_{x \to 0} \left(\frac{\sin \left(\frac{x}{2}\right)}{\frac{x}{2}}\right)^2 \cdot \left(\frac{x}{4}\right)$$

$$= 2 \cdot 1 \cdot \left(\frac{0}{4}\right)$$

$$= 0$$

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VSAQ-17 : Evaluate Lt x→0 1 – cos2mx/sin²nx

$$Lt_{x \to 0} \frac{2 \sin^2(mx)}{\sin^2(nx)}$$

$$= 2 \cdot Lt_{x \to 0} \left( \frac{\sin(mx)}{mx} \right)^2 \cdot \left( \frac{nx}{\sin(nx)} \right)^2 \cdot \frac{m^2 x^2}{n^2 x^2}$$

$$= 2 \cdot (1)^2 \cdot (1)^2 \cdot \frac{m^2}{n^2}$$

$$= \frac{2m^2}{n^2}$$

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VSAQ-18 : Find Lt x→0 a^x – 1/b^x – 1

$$Lt_{x \to 0} \frac{ax – 1}{bx – 1}$$

$$= Lt_{x \to 0} \frac{\frac{ax – 1}{x}}{\frac{bx – 1}{x}}$$

$$= \frac{Lt_{x \to 0} (ax – 1)/x}{Lt_{x \to 0} (bx – 1)/x}$$

$$= \frac{\log_e a}{\log_e b}$$

$$= \log_b a$$

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VSAQ-19 : Evaluate Lt x→0 e^7x – 1/x

$$Lt_{x \to 0} \frac{e^{7x} – 1}{x}$$

$$= Lt_{7x \to 0} \frac{e^{7x} – 1}{7x} \cdot 7$$

$$= \left( Lt_{u \to 0} \frac{e^u – 1}{u} \right) \cdot 7$$

$$= 1 \cdot 7$$

$$= 7$$

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VSAQ-20 : Evaluate Lt x→0 e^3x – 1/x

$$Lt_{x \to 0} \frac{e^{3x} – 1}{x}$$

$$= Lt_{3x \to 0} \frac{e^{3x} – 1}{3x} \cdot 3$$

$$= 1 \cdot 3$$

$$= 3$$

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VSAQ-21 : Find Lt x→0 e^3+x – e³/x

$$Lt_{x \to 0} \frac{e^{3+x} – e^3}{x}$$

$$= Lt_{x \to 0} \frac{e^3 (e^x – 1)}{x}$$

$$= e^3 \cdot Lt_{x \to 0} \frac{e^x – 1}{x}$$

$$= e^3 \cdot 1$$

$$= e^3$$

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VSAQ-22 : Evaluate Lt x→0 e^sinx – 1/x

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

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

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

$$= 1 \cdot 1$$

$$= 1$$

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VSAQ-23 : Evaluate Lt x→0 log(1 + 5x)/x

$$Lt_{x \to 0} \frac{\log(1 + 5x)}{x}$$

$$= Lt_{5x \to 0} \frac{\log(1 + 5x)}{5x} \cdot 5$$

$$= \left( Lt_{u \to 0} \frac{\log(1 + u)}{u} \right) \cdot 5$$

$$= 1 \cdot 5$$

$$= 5$$

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VSAQ-24 : Evaluate Lt x→1 log_ex/x – 1

$$\text{Put } y = x – 1$$

$$\text{Then } x \to 1$$

$$\Rightarrow x – 1 \to 0$$

$$\Rightarrow y \to 0$$

$$Lt_{x \to 1} \frac{\log_e x}{x – 1} = Lt_{y \to 0} \frac{\log(1 + y)}{y}$$

$$= 1$$

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VSAQ-25 : Compute Lt x→∞ x² + 5x + 2/2x² – 5x + 1

$$\text{Dividing the numerator and denominator by } x^2 \text{, the highest power of } x, \text{ we have}$$

$$Lt_{x \to \infty} \frac{x^2 + 5x + 2}{2x^2 – 5x + 1}$$

$$= Lt_{x \to \infty} \frac{\frac{x^2}{x^2} + \frac{5x}{x^2} + \frac{2}{x^2}}{\frac{2x^2}{x^2} – \frac{5x}{x^2} + \frac{1}{x^2}}$$

$$= Lt_{1/x \to 0} \frac{1 + \frac{5}{x} + \frac{2}{x^2}}{2 – \frac{5}{x} + \frac{1}{x^2}}$$

$$= \frac{1 + 0 + 0}{2 – 0 + 0}$$

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

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VSAQ-26 : Evaluate Lt x→∞ 11x³ – 3x + 4/13x³ – 5x² – 7

Taking x³, the highest power of x as common factor in the numerator and denominator

$$Lt_{x \to \infty} \frac{11x^3 – 3x + 4}{13x^3 – 5x^2 – 7}$$

$$= Lt_{x \to \infty} \frac{x^3(11 – \frac{3}{x^2} + \frac{4}{x^3})}{x^3(13 – \frac{5}{x} – \frac{7}{x^3})}$$

$$= Lt_{1/x \to 0} \frac{11 – \frac{3}{x^2} + \frac{4}{x^3}}{13 – \frac{5}{x} – \frac{7}{x^3}}$$

$$= \frac{11 – 0 + 0}{13 – 0 – 0}$$

$$= \frac{11}{13}$$

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VSAQ-27 : Find Lt x→∞ 8|x| + 3x/3|x| – 2x

$$\text{If } x \to \infty \text{ then } x > 0$$

$$\text{Hence } |x| = x$$

$$Lt_{x \to \infty} \frac{8|x| + 3x}{3|x| – 2x}$$

$$= Lt_{x \to \infty} \frac{8x + 3x}{3x – 2x}$$

$$= Lt_{x \to \infty} \frac{11x}{x}$$

$$= Lt_{x \to \infty} 11$$

$$= 11$$

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VSAQ-28 : Compute Lt x→2 x² + 2x – 1/x² – 4x + 4

$$Lt_{x \to 2} \frac{x^2 + 2x – 1}{x^2 – 4x + 4}$$

$$= Lt_{x \to 2} \frac{x^2 + 2x – 1}{(x – 2)^2}$$

$$= \frac{4 + 4 – 1}{(2 – 2)^2}$$

$$= \frac{7}{0}$$

$$= \infty$$

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VSAQ-29 : Evaluate Lt x→3 x² + 3x + 2/x² – 6x + 9

$$Lt_{x \to 3} \frac{x^2 + 3x + 2}{x^2 – 6x + 9}$$

$$= Lt_{x \to 3} \frac{x^2 + 3x + 2}{(x – 3)^2}$$

$$= \frac{9 + 9 + 2}{(3 – 3)^2}$$

$$= \frac{20}{0}$$

$$= \infty$$

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VSAQ-30 : Show that Lt x→0+ (2|x|/x + x + 1) = 3

$$\text{When } x \to 0^+ \text{ then } x > 0$$

$$\text{|x| = x}$$

$$Lt_{x \to 0^+} \left( \frac{2|x|}{x} + x + 1 \right)$$

$$= Lt_{x \to 0^+} \left( \frac{2x}{x} + x + 1 \right)$$

$$= 2 + 0 + 1$$

$$= 3$$

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VSAQ-31 : Show that Lt x→2- |x – 2|/x – 2 = -1

$$\text{When } x \to 2^- \text{ then } x < 2$$

$$\Rightarrow (x – 2) < 0$$

$$\Rightarrow |x – 2| = -(x – 2)$$

$$Lt_{x \to 2^-} \frac{|x – 2|}{x – 2}$$

$$= Lt_{x \to 2^-} \frac{-(x – 2)}{x – 2}$$

$$= -1$$