CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 08

CBSE Class 11 Mathematics Worksheet - Limits and Derivatives. The questions in the worksheets have been specifically designed by best teachers so that the students can practise them to clear their concepts and get better marks in tests and examinations. Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Evaluate: \( \lim_{x\to0} \left( \frac{\text{cosec } x - \text{cot } x}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{\text{cosec } x - \text{cot } x}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{1 - \cos x}{x \sin x} \right) \)
\( = \lim_{x\to0} \left( \frac{2 \sin^2\left(\frac{x}{2}\right)}{x \sin x} \right) \)
\( = \lim_{x\to0} \left( \frac{ \frac{2 \sin^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}} \times \frac{x^2}{4} }{ \frac{x \sin x}{x} \times x } \right) \)
\( = \frac{2}{4} \frac{\lim_{x\to0} \left( \frac{\sin^2\left(\frac{x}{2}\right)}{\frac{x^2}{4}} \right)}{\lim_{x\to0} \left( \frac{\sin x}{x} \right)} \)
\( = \frac{1}{2} \times \frac{(1)}{(1)} \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = \frac{1}{2} \) ans.

Question. Evaluate: \( \lim_{x\to\frac{\pi}{2}} \left( \frac{1+\cos(2x)}{(\pi-2x)^2} \right) \)
Answer: We have : \( \lim_{x\to\frac{\pi}{2}} \left( \frac{1+\cos(2x)}{(\pi-2x)^2} \right) \)
Put \( x = \frac{\pi}{2} + h \) and \( h \to 0 \)
\( \therefore \lim_{h\to0} \left( \frac{1 + \cos \left(2 \left(\frac{\pi}{2} + h\right)\right)}{\left(\pi - 2 \left(\frac{\pi}{2} + h\right)\right)^2} \right) \)
\( = \lim_{h\to0} \left( \frac{1 + \cos(\pi + 2h)}{(\pi - \pi - 2h)^2} \right) \)
\( = \lim_{h\to0} \left( \frac{1-\cos(2h)}{4h^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2 \sin^2 h}{4h^2} \right) \)
\( = \frac{1}{2} \lim_{h\to0} \left( \frac{\sin^2 h}{h^2} \right) \)
\( = \frac{1}{2} (1)^2 \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = \frac{1}{2} \) ans.

Question. Evaluate: \( \lim_{x\to\frac{\pi}{6}} \left( \frac{2-\sqrt{3} \cos x-\sin x}{(6x-\pi)^2} \right) \)
Answer: We have \( \lim_{x\to\frac{\pi}{6}} \left( \frac{2-\sqrt{3} \cos x-\sin x}{(6x-\pi)^2} \right) \)
Put \( x = \frac{\pi}{6} + h \) and \( h \to 0 \)
\( = \lim_{h\to0} \left( \frac{2 - \sqrt{3} \cos \left(\frac{\pi}{6} + h\right) - \sin \left(\frac{\pi}{6} + h\right)}{\left(6 \left(\frac{\pi}{6} + h\right) - \pi\right)^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2 - \sqrt{3} \left(\cos\frac{\pi}{6} \cos h - \sin\frac{\pi}{6} \sin h\right) - \left(\sin\frac{\pi}{6} \cos h + \cos\frac{\pi}{6} \sin h\right)}{(\pi + 6h - \pi)^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2 - \sqrt{3} \left(\frac{\sqrt{3}}{2} \cos h - \frac{1}{2} \sin h\right) - \left(\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h\right)}{36h^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2 - \frac{3}{2} \cos h + \frac{\sqrt{3}}{2} \sin h - \frac{1}{2} \cos h - \frac{\sqrt{3}}{2} \sin h}{36h^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2 - 2 \cos h}{36h^2} \right) \)
\( = \lim_{h\to0} \left( \frac{2(1 - \cos h)}{36h^2} \right) \)
\( = \frac{1}{18} \lim_{h\to0} \left( \frac{2 \sin^2 \left(\frac{h}{2}\right)}{\frac{h^2}{4} \times 4} \right) \)
\( = \frac{1}{36} \lim_{h\to0} \left( \frac{\sin^2 \left(\frac{h}{2}\right)}{\frac{h^2}{4}} \right) \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = \frac{1}{36} \times (1)^2 = \frac{1}{36} \) ans.

Question. Evaluate: \( \lim_{x\to\pi} \left( \frac{\sin(3x)-3 \sin x}{(\pi-x)^3} \right) \)
Answer: We have \( \lim_{x\to\pi} \left( \frac{\sin(3x)-3 \sin x}{(\pi-x)^3} \right) \)
\( = \lim_{x\to\pi} \left( \frac{3 \sin x-4 \sin^3 x-3 \sin x}{(\pi-x)^3} \right) \)
\( = \lim_{x\to\pi} \left( \frac{-4 \sin^3 x}{(\pi-x)^3} \right) \)
Put \( x = \pi + h \) and \( h \to 0 \)
\( = \lim_{h\to0} \left( \frac{-4 \sin^3(\pi + h)}{(\pi - \pi - h)^3} \right) \)
\( = \lim_{h\to0} \left( \frac{4 \sin^3 h}{-h^3} \right) \) \quad \( \{\sin(\pi + \theta) = - \sin \theta\} \)
\( = -4 \lim_{h\to0} \left( \frac{\sin^3 h}{h^3} \right) \)
\( = -4(1)^3 = -4 \) ans. \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)

Question. Evaluate: \( \lim_{x\to\frac{\pi}{2}} \left( \frac{\cot x-\cos x}{(\pi-2x)^3} \right) \)
Answer: We have \( \lim_{x\to\frac{\pi}{2}} \left( \frac{\cot x-\cos x}{(\pi-2x)^3} \right) \)
Put \( x = \frac{\pi}{2} + h \) and \( h \to 0 \)
\( = \lim_{h\to0} \left( \frac{\cot \left(\frac{\pi}{2} + h\right) - \cos \left(\frac{\pi}{2} + h\right)}{\left(\pi - 2 \left(\frac{\pi}{2} + h\right)\right)^3} \right) \)
\( = \lim_{h\to0} \left( \frac{- \tan h+\sin h}{(\pi-\pi-2h)^3} \right) \)
\( = \lim_{h\to0} \left( \frac{- \tan h+\sin h}{-8h^3} \right) \)
\( = \lim_{h\to0} \left( \frac{\tan h - \sin h}{8h^3} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\frac{\sin h}{\cos h} - \sin h}{h^3} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h - \sin h \cdot \cos h}{h^3 \cdot \cos h} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h (1 - \cos h)}{h^3 \cdot \cos h} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h \cdot 2 \sin^2 \left(\frac{h}{2}\right)}{h^3 \cdot \cos h} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h}{h} \cdot \frac{2 \sin^2 \left(\frac{h}{2}\right)}{h^2} \cdot \frac{1}{\cos h} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h}{h} \cdot \frac{2 \sin^2 \left(\frac{h}{2}\right)}{\frac{h^2}{4} \times 4} \cdot \frac{1}{\cos h} \right) \)
\( = \frac{1}{8} \lim_{h\to0} \left( \frac{\sin h}{h} \right) \times \frac{2}{4} \cdot \lim_{h\to0} \left( \frac{\sin^2 \left(\frac{h}{2}\right)}{\frac{h^2}{4}} \right) \times \lim_{h\to0} \left( \frac{1}{\cos h} \right) \)
\( = \frac{1}{8} \times 1 \times \frac{1}{2} \times 1^2 \times 1 \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = \frac{1}{16} \) ans.

Question. Evaluate: \( \lim_{x\to\frac{\pi}{2}} \left( \frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos^2 x} \right) \)
Answer: We have \( \lim_{x\to\frac{\pi}{2}} \left( \frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos^2 x} \right) \)
Rationalize
\( = \lim_{x\to\frac{\pi}{2}} \left[ \frac{(\sqrt{2}-\sqrt{1+\sin x})(\sqrt{2}+\sqrt{1+\sin x})}{\cos^2 x(\sqrt{2}+\sqrt{1+\sin x})} \right] \)
\( = \lim_{x\to\frac{\pi}{2}} \left[ \frac{2-1-\sin x}{\cos^2 x(\sqrt{2}+\sqrt{1+\sin x})} \right] \)
\( = \lim_{x\to\frac{\pi}{2}} \left[ \frac{1-\sin x}{\cos^2 x(\sqrt{2}+\sqrt{1+\sin x})} \right] \)
Put \( x = \frac{\pi}{2} + h \) and \( h \to 0 \)
\( = \lim_{h\to0} \left[ \frac{1 - \sin \left(\frac{\pi}{2} + h\right)}{\cos^2 \left(\frac{\pi}{2} + h\right)} \right] \times \lim_{x\to\frac{\pi}{2}} \left[ \frac{1}{\sqrt{2} + \sqrt{1 + \sin x}} \right] \)
\( = \lim_{h\to0} \left( \frac{1 - \cos h}{\sin^2 h} \right) \times \frac{1}{\sqrt{2} + \sqrt{2}} \)
\( = \lim_{h\to0} \left( \frac{2 \sin^2 \left(\frac{h}{2}\right)}{\sin^2 h} \right) \times \frac{1}{2\sqrt{2}} \)
\( = \lim_{h\to0} \left( \frac{\frac{2 \sin^2 \left(\frac{h}{2}\right)}{\frac{h^2}{4}} \times \frac{h^2}{4}}{\frac{\sin^2 h}{h^2} \times h^2} \right) \times \frac{1}{2\sqrt{2}} \)
\( = \frac{2}{4} \frac{\lim_{h\to0} \left( \frac{\sin^2\left(\frac{h}{2}\right)}{\frac{h^2}{4}} \right)}{\lim_{h\to0} \left( \frac{\sin^2 h}{h^2} \right)} \times \frac{1}{2\sqrt{2}} \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = \frac{1}{2} \times \frac{1}{1} \times \frac{1}{2\sqrt{2}} \)
\( = \frac{1}{4\sqrt{2}} \) ans.

Question. Evaluate: \( \lim_{x\to0} \left( \frac{1-\cos(2x)}{\cos(2x)-\cos(8x)} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{1-\cos(2x)}{\cos(2x)-\cos(8x)} \right) \)
\( = \lim_{x\to0} \left[ \frac{2 \sin^2 𝑥}{-2 \sin(5𝑥). \sin(-3𝑥)} \right] \)
\( = \lim_{x\to0} \left[ \frac{\sin^2 x}{x^2 \times x^2}{\frac{\sin(5x)}{5x} \times \frac{\sin(3x)}{3x} \times 15} \right] \)
\( = \frac{\lim_{x\to0} \frac{\sin^2 𝑥}{𝑥^2}}{15 \times \lim_{x\to0} \frac{\sin(5𝑥)}{5𝑥} \times \lim_{x\to0} \frac{\sin(3𝑥)}{3𝑥}} \)
\( = \frac{1^2}{15(1)(1)} = \frac{1}{15} \) ans. \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)

Question. Evaluate: \( \lim_{x\to a} \left( \frac{\sin 𝑥-\sin 𝑎}{\sqrt{𝑥}-\sqrt{𝑎}} \right) \)
Answer: We have \( \lim_{x\to a} \left( \frac{\sin 𝑥-\sin 𝑎}{\sqrt{𝑥}-\sqrt{𝑎}} \right) \)
Rationalize
\( = \lim_{x\to a} \left( \frac{(\sin 𝑥-\sin 𝑎)(\sqrt{𝑥}+\sqrt{𝑎})}{𝑥-𝑎} \right) \)
Put \( 𝑥 = 𝑎 + ℎ \) and \( ℎ \to 0 \)
\( = \lim_{ℎ\to0} \left( \frac{(\sin(𝑎+ℎ)-\sin 𝑎)(\sqrt{𝑎+ℎ}+\sqrt{𝑎})}{𝑎+ℎ-𝑎} \right) \)
\( = \lim_{ℎ\to0} \left( \frac{2 \cos\left(\frac{2𝑎+ℎ}{2}\right) \cdot \sin\left(\frac{ℎ}{2}\right) \cdot (\sqrt{𝑎+ℎ}+\sqrt{𝑎})}{ℎ} \right) \)
\( = \lim_{ℎ\to0} \left( \frac{2 \cos\left(\frac{2𝑎+ℎ}{2}\right) \cdot \sin\left(\frac{ℎ}{2}\right) \cdot (\sqrt{𝑎+ℎ}+\sqrt{𝑎})}{\frac{ℎ}{2} \times 2} \right) \)
\( = \lim_{ℎ\to0} \left( \frac{\sin\left(\frac{ℎ}{2}\right)}{\frac{ℎ}{2}} \right) \times \lim_{ℎ\to0} \left( \cos \left(\frac{2𝑎+ℎ}{2}\right) \cdot (\sqrt{𝑎+ℎ} - \sqrt{𝑎}) \right) \)
\( = 1 \times \cos(a)(\sqrt{a} + \sqrt{a}) \) \quad \( \left\{\lim_{x\to0} \left( \frac{\sin x}{x} \right) = 1\right\} \)
\( = 2\sqrt{a} \cos a \) ans.

Question. Evaluate \( \lim_{x\to0} \left( \frac{a^x-b^x}{x} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{a^x-b^x}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x-b^x-1+1}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{(a^x-1)-(b^x-1)}{x} \right) \)
\( = \lim_{x\to0} \left( \frac{a^x-1}{x} \right) - \lim_{x\to0} \left( \frac{b^x-1}{x} \right) \)
\( = \log a - \log b \)
\( = \log \left(\frac{a}{b}\right) \) ans. \quad \( \left\{\lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a\right\} \)

Question. Evaluate \( \lim_{x\to0} \left( \frac{2^x-1}{\sqrt{1+x}-1} \right) \)
Answer: We have \( \lim_{x\to0} \left( \frac{2^x-1}{\sqrt{1+x}-1} \right) \)
Rationalize
\( = \lim_{x\to0} \left( \frac{2^x-1}{\sqrt{1+x}-1} \times \frac{(\sqrt{1+x}+1)}{(\sqrt{1+x}+1)} \right) \)
\( = \lim_{x\to0} \left( \frac{(2^x-1)(\sqrt{1+x}+1)}{1+x-1} \right) \)
\( = \lim_{x\to0} \left( \frac{(2^x-1)}{1+x-1} \right) \times \lim_{x\to0} (\sqrt{1 + 𝑥} + 1) \)
\( = \log 2 \times (1 + 1) \)
\( = 2 \log 2 \) ans. \quad \( \left\{\lim_{x\to0} \left( \frac{a^x-1}{x} \right) = \log a\right\} \)