CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 05

TYPE: 5 TRIGO LIMITS

Question 1. Evaluate: \(\lim_{x \to 0} \left(\frac{\sec x - \cot x}{x}\right)\)
Answer: We have \(\lim_{x \to 0} \left(\frac{\sec x - \cot x}{x}\right)\)
\(= \lim_{x \to 0} \left( \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{1 - \cos x}{x \sin x} \right)\)
\(= \lim_{x \to 0} \left( \frac{2 \sin^2 \left(\frac{x}{2}\right)}{x \sin x} \right)\)
Dividing numerator and denominator appropriately to use standard limits:
\(= \lim_{x \to 0} \left( \frac{2 \sin^2 \left(\frac{x}{2}\right)}{\frac{x^2}{4} \times 4} \times \frac{x}{\sin x \cdot x} \times x \right)\) - wait, let us simplify directly:
\(= \frac{1}{2} \times \lim_{x \to 0} \left[ \frac{\sin^2 \left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \right] \div \lim_{x \to 0} \left(\frac{\sin x}{x}\right)\)
Using standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{1}{2} \times (1)^2 \times 1 = \frac{1}{2}\)
In simple words: Convert secant and cotangent into sine and cosine, use the half-angle formula for \(1 - \cos x\), and apply the standard trigonometric limit.

Exam Tip: Always express trigonometric ratios in terms of \(\sin x\) and \(\cos x\) when stuck, and remember the standard limit \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

TYPE: 6 TRIGO LIMITS

Question 2. 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 \(x = \frac{\pi}{2} + h\) such that as \(x \to \frac{\pi}{2}\), \(h \to 0\).
\(= \lim_{h \to 0} \left( \frac{1 + \cos\left(2\left(\frac{\pi}{2} + h\right)\right)}{\left(\pi - 2\left(\frac{\pi}{2} + h\right)\right)^2} \right)\)
\(= \lim_{h \to 0} \left( \frac{1 + \cos(\pi + 2h)}{(\pi - \pi - 2h)^2} \right)\)
\(= \lim_{h \to 0} \left( \frac{1 - \cos(2h)}{4h^2} \right)\)
\(= \lim_{h \to 0} \left( \frac{2 \sin^2 h}{4h^2} \right)\)
\(= \frac{1}{2} \lim_{h \to 0} \left( \frac{\sin h}{h} \right)^2\)
\(= \frac{1}{2}(1)^2 = \frac{1}{2}\)
In simple words: Substitute \(x = \frac{\pi}{2} + h\) to shift the limit to zero, simplify using trigonometric identities, and apply the standard limit.

Exam Tip: Substitution of variables (like putting \(x = a + h\)) is a powerful tool to evaluate limits approaching a specific angle like \(\frac{\pi}{2}\) or \(\frac{\pi}{6}\).

Question 3. 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 \(x = \frac{\pi}{6} + h\) where \(h \to 0\).
\(= \lim_{h \to 0} \left( \frac{2 - \sqrt{3}\cos\left(\frac{\pi}{6} + h\right) - \sin\left(\frac{\pi}{6} + h\right)}{\left(6\left(\frac{\pi}{6} + h\right) - \pi\right)^2} \right)\)
Expanding using angle sum formulas: trigonometric expansion gives
\(= \lim_{h \to 0} \left( \frac{2 - 2\cos h}{36h^2} \right)\)
\(= \lim_{h \to 0} \left( \frac{2(1 - \cos h)}{36h^2} \right)\)
\(= \frac{1}{18} \lim_{h \to 0} \left( \frac{2\sin^2\left(\frac{h}{2}\right)}{h^2} \right)\)
\(= \frac{1}{36} \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right)^2\)
\(= \frac{1}{36} \times (1)^2 = \frac{1}{36}\)
In simple words: Shift the limit variable using substitution, expand the compound angles, and simplify terms until standard limits can be applied.

Exam Tip: Be careful with coefficient expansions when substituting linear multiples of \(x\) like \(6x - \pi\). Factor out constants carefully.

Question 5. Evaluate: \(\lim_{x \to \pi} \left(\frac{\sin(3x) - 3\sin x}{(\pi - x)^3}\right)\)
Answer: We have \(\lim_{x \to \pi} \left(\frac{\sin(3x) - 3\sin x}{(\pi - x)^3}\right)\)
Using the identity \(\sin(3x) = 3\sin x - 4\sin^3 x\):
\(= \lim_{x \to \pi} \left( \frac{3\sin x - 4\sin^3 x - 3\sin x}{(\pi - x)^3} \right)\)
\(= \lim_{x \to \pi} \left( \frac{-4\sin^3 x}{(\pi - x)^3} \right)\)
Put \(x = \pi + h\) where \(h \to 0\). Since \(\sin(\pi + h) = -\sin h\):
\(= \lim_{h \to 0} \left( \frac{-4(-\sin h)^3}{(-h)^3} \right)\)
\(= \lim_{h \to 0} \left( \frac{4\sin^3 h}{-h^3} \right)\)
\(= -4 \lim_{h \to 0} \left( \frac{\sin h}{h} \right)^3\)
\(= -4(1)^3 = -4\)
In simple words: Use triple-angle identity for sine to eliminate linear sine terms, then substitute \(x = \pi + h\) to solve.

Exam Tip: Memorize triple angle formulas like \(\sin(3x) = 3\sin x - 4\sin^3 x\) as they frequently appear in limit problems involving \(\pi\).

Question 6. 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 \(x = \frac{\pi}{2} + h\) where \(h \to 0\):
\(= \lim_{h \to 0} \left( \frac{\cot\left(\frac{\pi}{2} + h\right) - \cos\left(\frac{\pi}{2} + h\right)}{\left(\pi - 2\left(\frac{\pi}{2} + h\right)\right)^3} \right)\)
\(= \lim_{h \to 0} \left( \frac{-\tan h + \sin h}{-8h^3} \right)\)
\(= \lim_{h \to 0} \left( \frac{\tan h - \sin h}{8h^3} \right)\)
\(= \lim_{h \to 0} \left( \frac{\sin h (1 - \cos h)}{8h^3 \cos h} \right)\)
\(= \frac{1}{8} \lim_{h \to 0} \left[ \left(\frac{\sin h}{h}\right) \cdot \left(\frac{2\sin^2\left(\frac{h}{2}\right)}{h^2}\right) \cdot \left(\frac{1}{\cos h}\right) \right]\)
\(= \frac{1}{8} \times 1 \times \frac{1}{2} \times 1 = \frac{1}{16}\)
In simple words: Substitute \(x = \frac{\pi}{2} + h\), convert \(\tan h\) to sine and cosine, use \(1 - \cos h = 2\sin^2\left(\frac{h}{2}\right)\), and evaluate.

Exam Tip: Group terms into standard limits like \(\frac{\sin h}{h}\) and \(\frac{1}{\cos h}\) to evaluate products easily.

Question 7. 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 the numerator:
\(= \lim_{x \to \frac{\pi}{2}} \left( \frac{2 - 1 - \sin x}{\cos^2 x (\sqrt{2} + \sqrt{1 + \sin x})} \right)\)
\(= \lim_{x \to \frac{\pi}{2}} \left( \frac{1 - \sin x}{(1 - \sin^2 x)(\sqrt{2} + \sqrt{1 + \sin x})} \right)\)
\(= \lim_{x \to \frac{\pi}{2}} \left( \frac{1}{(1 + \sin x)(\sqrt{2} + \sqrt{1 + \sin x})} \right)\)
Substitute \(x = \frac{\pi}{2} + h\) or evaluate directly since the indeterminate form is resolved:
\(= \frac{1}{(1 + 1)(\sqrt{2} + \sqrt{1 + 1})} = \frac{1}{2 \times 2\sqrt{2}} = \frac{1}{4\sqrt{2}}\) (or rationalized as \(\frac{\sqrt{2}}{8}\)).
In simple words: Rationalize the numerator containing square roots, cancel common factors from \(\cos^2 x = 1 - \sin^2 x\), and substitute the limit directly.

Exam Tip: Always look for rationalization opportunities when square roots appear in limits.

Question 8. Evaluate: \(\lim_{x \to 0} \left(\frac{1 - \cos(2x)}{\cos(2x) - \cos(8x)}\right)\)
Answer: We have \(\lim_{x \to 0} \left(\frac{1 - \cos(2x)}{\cos(2x) - \cos(8x)}\right)\)
Using trigonometric formulas \(1 - \cos(2x) = 2\sin^2 x\) and \(\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{2\sin^2 x}{-2\sin(5x)\sin(-3x)} \right)\)
\(= \lim_{x \to 0} \left( \frac{\sin^2 x}{\sin(5x)\sin(3x)} \right)\)
Divide numerator and denominator by \(x^2\):
\(= \lim_{x \to 0} \left( \frac{\left(\frac{\sin x}{x}\right)^2}{\left(\frac{\sin(5x)}{5x} \times 5\right) \left(\frac{\sin(3x)}{3x} \times 3\right)} \right)\)
\(= \frac{(1)^2}{5 \times 3} = \frac{1}{15}\)
In simple words: Transform products and differences using sum-to-product formulas, then divide by \(x^2\) to form standard limits.

Question 9. Evaluate: \(\lim_{x \to a} \left(\frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}\right)\)
Answer: We have \(\lim_{x \to a} \left(\frac{\sin x - \sin a}{\sqrt{x} - \sqrt{a}}\right)\)
Using sum-to-product formula for \(\sin x - \sin a\) and rationalizing or substituting \(x = a + h\):
Put \(x = a + h\) where \(h \to 0\):
\(= \lim_{h \to 0} \left( \frac{\sin(a + h) - \sin a}{\sqrt{a + h} - \sqrt{a}} \right)\)
\(= \lim_{h \to 0} \left( \frac{2\cos\left(\frac{2a + h}{2}\right)\sin\left(\frac{h}{2}\right)}{\frac{(a + h) - a}{\sqrt{a + h} + \sqrt{a}}} \right)\) - wait, rationalizing denominator directly:
\(= \lim_{h \to 0} \left[ \frac{2\cos\left(\frac{2a+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} \times (\sqrt{a+h} + \sqrt{a}) \right]\)
\(= \lim_{h \to 0} \left[ \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right] \times \lim_{h \to 0} \left[ \cos\left(\frac{2a+h}{2}\right)(\sqrt{a+h} + \sqrt{a}) \right]\)
\(= 1 \times \cos a (\sqrt{a} + \sqrt{a}) = 2\sqrt{a}\cos a\)
In simple words: Apply the trigonometric difference formula for sines and simplify with standard limits.

Exam Tip: Combining substitution with standard trigonometric limits makes complex algebraic-trigonometric hybrids solvable.

TYPE: 7

Question 10. Evaluate: \(\lim_{x \to 0} \left(\frac{a^x - b^x}{x}\right)\)
Answer: We have \(\lim_{x \to 0} \left(\frac{a^x - b^x}{x}\right)\)
Add and subtract 1 in the numerator:
\(= \lim_{x \to 0} \left( \frac{a^x - b^x - 1 + 1}{x} \right)\)
\(= \lim_{x \to 0} \left( \frac{(a^x - 1) - (b^x - 1)}{x} \right)\)
\(= \lim_{x \to 0} \left(\frac{a^x - 1}{x}\right) - \lim_{x \to 0} \left(\frac{b^x - 1}{x}\right)\)
Using the standard limit formula \(\lim_{x \to 0} \frac{k^x - 1}{x} = \log k\):
\(= \log a - \log b = \log\left(\frac{a}{b}\right)\)
In simple words: Introduce \(-1\) and \(+1\) to split the expression into two standard exponential limit forms.

Exam Tip: Remember the standard exponential limit: \(\lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a\).