CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 06

Exam Tip: Always divide numerator and denominator by the highest independent variable power in the denominator to convert terms into standard limit forms.

Question 2. Evaluate: \(\lim_{x \to 0} \left( \frac{1 - \cos(mx)}{1 - \cos(nx)} \right)\)
Answer: Using the trigonometric identity \(1 - \cos(2\theta) = 2\sin^2(\theta)\):
\(= \lim_{x \to 0} \left( \frac{2\sin^2\left(\frac{mx}{2}\right)}{2\sin^2\left(\frac{nx}{2}\right)} \right)\)
Rearranging terms to match the standard limit form \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \lim_{x \to 0} \left( \frac{\frac{\sin^2\left(\frac{mx}{2}\right)}{\left(\frac{mx}{2}\right)^2} \cdot \left(\frac{mx}{2}\right)^2}{\frac{\sin^2\left(\frac{nx}{2}\right)}{\left(\frac{nx}{2}\right)^2} \cdot \left(\frac{nx}{2}\right)^2} \right)\)
\(= \frac{m^2}{n^2} \frac{\lim_{x \to 0} \left(\frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2}{\lim_{x \to 0} \left(\frac{\sin\left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2}\)
\(= \frac{m^2(1)^2}{n^2(1)^2} = \frac{m^2}{n^2}\)
In simple words: Apply the half-angle formula for cosine and adjust the denominators to apply the standard limit for sine squared.

Exam Tip: Memorize the half-angle formula \(1 - \cos(A) = 2\sin^2\left(\frac{A}{2}\right)\) as it simplifies trigonometric limits immensely.

Question 3. Evaluate: \(\lim_{x \to 0} (\csc x - \cot x)\)
Answer: Express cosecant and cotangent in terms of sine and cosine:
\(= \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right)\)
\(= \lim_{x \to 0} \left( \frac{1 - \cos x}{\sin x} \right)\)
Substitute \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\) and \(\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)} \right)\)
\(= \lim_{x \to 0} \tan\left(\frac{x}{2}\right)\)
\(= \tan(0) = 0\)
In simple words: Convert cosecant and cotangent into sine and cosine fractions, simplify, and directly substitute the limit value.

Exam Tip: Converting trigonometric ratios into sine and cosine is a reliable fallback method when direct evaluation fails.

Question 4. Evaluate: \(\lim_{x \to 0} \left( \frac{x^3 \cot x}{1 - \cos x} \right)\)
Answer: Rewrite cotangent as \(\frac{\cos x}{\sin x}\) and use \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{x^3 \cdot \frac{\cos x}{\sin x}}{2\sin^2\left(\frac{x}{2}\right)} \right)\)
\(= \lim_{x \to 0} \left( \frac{x^3}{\sin x} \cdot \frac{1}{2\sin^2\left(\frac{x}{2}\right)} \cdot \cos x \right)\)
Adjusting terms for standard limits:
\(= \lim_{x \to 0} \left( \frac{x}{\sin x} \cdot \frac{x^2}{2\sin^2\left(\frac{x}{2}\right)} \cdot \cos x \right)\)
\(= \left( \lim_{x \to 0} \frac{x}{\sin x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{2 \left(\frac{\sin(x/2)}{x/2}\right)^2 \cdot \frac{1}{4}} \right) \cdot \left( \lim_{x \to 0} \cos x \right)\)
\(= 1 \cdot \left( \frac{1}{2 \cdot \frac{1}{4} \cdot (1)^2} \right) \cdot 1\)
\(= \frac{1}{\frac{1}{2}} = 2\)
In simple words: Break the expression into standard limit components for \(\frac{\sin x}{x}\) and \(\cos x\), then multiply the evaluated parts.

Exam Tip: Carefully manage coefficients like \(\frac{1}{4}\) when adjusting the angle inside squared sine terms.

Question 5. Evaluate: \(\lim_{x \to 0} \left( \frac{\sin(2x) + \sin(6x)}{\sin(5x) - \sin(3x)} \right)\)
Answer: Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \left( \frac{2 \left(\frac{\sin(2x)}{2x}\right) + 6 \left(\frac{\sin(6x)}{6x}\right)}{5 \left(\frac{\sin(5x)}{5x}\right) - 3 \left(\frac{\sin(3x)}{3x}\right)} \right)\)
Using \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{2(1) + 6(1)}{5(1) - 3(1)}\)
\(= \frac{2 + 6}{5 - 3}\)
\(= \frac{8}{2} = 4\)
In simple words: Multiply and divide each angle by its respective coefficient to form standard limits, then simplify the numerical fraction.

Exam Tip: This technique of multiplying and dividing by the angle coefficient is known as making the argument standard.

Question 6. Evaluate: \(\lim_{x \to 0} \left( \frac{\tan x - \sin x}{x^3} \right)\)
Answer: Substitute \(\tan x = \frac{\sin x}{\cos x}\):
\(= \lim_{x \to 0} \left( \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \right)\)
\(= \lim_{x \to 0} \left( \frac{\sin x(1 - \cos x)}{x^3 \cos x} \right)\)
Using \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{\sin x \cdot 2\sin^2\left(\frac{x}{2}\right)}{x^3 \cos x} \right)\)
\(= \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{2\sin^2\left(\frac{x}{2}\right)}{x^2} \cdot \frac{1}{\cos x} \right)\)
Adjusting the denominator for the half-angle limit:
\(= \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2 \cdot 4} \cdot \frac{2}{\cos x} \right)\)
\(= 1 \cdot \left(\frac{1}{4}\right) \cdot (1)^2 \cdot \left(\frac{2}{1}\right)\)
\(= \frac{2}{4} = \frac{1}{2}\)
In simple words: Factor out \(\sin x\), convert \(1 - \cos x\), and split into standard limits for \(\sin x\) and \(\tan x\).

Exam Tip: Keep track of constant multipliers like \(\frac{1}{4}\) when adjusting squared denominator terms.

Question 7. Evaluate: \(\lim_{x \to 0} \left( \frac{\tan x - \sin x}{\sin^3 x} \right)\)
Answer: Rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\):
\(= \lim_{x \to 0} \left( \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3 x} \right)\)
\(= \lim_{x \to 0} \left( \frac{\sin x(1 - \cos x)}{\sin^3 x \cos x} \right)\)
\(= \lim_{x \to 0} \left( \frac{1 - \cos x}{\sin^2 x \cos x} \right)\)
Using \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{2\sin^2\left(\frac{x}{2}\right)}{\sin^2 x \cos x} \right)\)
Adjusting limits:
\(= \lim_{x \to 0} \left( \frac{2\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2 \cdot 4} \cdot \frac{x^2}{\sin^2 x} \cdot \frac{1}{\cos x} \right)\)
\(= 2 \cdot \left(\frac{1}{4}\right) \cdot (1)^2 \cdot (1)^2 \cdot \left(\frac{1}{1}\right)\)
\(= \frac{2}{4} = \frac{1}{2}\)
In simple words: Cancel common \(\sin x\) terms from numerator and denominator, then apply the half-angle substitution for \(1 - \cos x\).

Exam Tip: Simplifying algebraic fractions before applying limit theorems reduces calculation errors.

Question 8. Evaluate: \(\lim_{x \to 0} \left( \frac{\sec(4x) - \sec(2x)}{\sec(3x) - \sec x} \right)\)
Answer: Convert secants to cosines:
\(= \lim_{x \to 0} \left( \frac{\frac{1}{\cos(4x)} - \frac{1}{\cos(2x)}}{\frac{1}{\cos(3x)} - \frac{1}{\cos x}} \right)\)
\(= \lim_{x \to 0} \left( \frac{\cos(2x) - \cos(4x)}{\cos(3x) - \cos x} \cdot \frac{\cos(3x)\cos x}{\cos(4x)\cos(2x)} \right)\)
Using the sum-to-product formula \(\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\):
\(= \lim_{x \to 0} \left( \frac{-2\sin(3x)\sin(-x)}{-2\sin(2x)\sin(-x)} \cdot \frac{\cos(3x)\cos x}{\cos(4x)\cos(2x)} \right)\)
\(= \lim_{x \to 0} \left( \frac{\sin(3x)}{\sin(2x)} \cdot \frac{\cos(3x)\cos x}{\cos(4x)\cos(2x)} \right)\)
Dividing numerator and denominator of the sine term by \(x\):
\(= \frac{3 \lim_{x \to 0} \left(\frac{\sin(3x)}{3x}\right)}{2 \lim_{x \to 0} \left(\frac{\sin(2x)}{2x}\right)} \cdot \lim_{x \to 0} \left( \frac{\cos(3x)\cos x}{\cos(4x)\cos(2x)} \right)\)
\(= \frac{3(1)}{2(1)} \cdot 1 = \frac{3}{2}\)
In simple words: Express secant as cosine, apply trigonometric factorization formulas, and split into standard limit forms.

Question 9. Evaluate:

Answer: Here, the variable is \(y\), and \(x\) is treated as a constant. Rewriting the expression:
\(= \lim_{y \to 0} \left( \frac{x\sec(x+y) + y\sec(x+y) - x\sec x}{y} \right)\)
\(= \lim_{y \to 0} \left( x \left( \frac{\sec(x+y) - \sec x}{y} \right) + \sec(x+y) \right)\)
Converting secant to cosine:
\(= \lim_{y \to 0} \left( x \left( \frac{\frac{1}{\cos(x+y)} - \frac{1}{\cos x}}{y} \right) \right) + \sec x\)
\(= \lim_{y \to 0} \left( x \left( \frac{\cos x - \cos(x+y)}{y \cos(x+y)\cos x} \right) \right) + \sec x\)
Using the formula for \(\cos A - \cos B\):
\(= \lim_{y \to 0} \left( x \left( \frac{-2\sin\left(\frac{2x+y}{2}\right)\sin\left(\frac{-y}{2}\right)}{y \cos(x+y)\cos x} \right) \right) + \sec x\)
\(= \lim_{y \to 0} \left( \frac{\sin(y/2)}{y/2} \right) \cdot \lim_{y \to 0} \left( \frac{x \sin\left(\frac{2x+y}{2}\right)}{\cos(x+y)\cos x} \right) + \sec x\)
\(= 1 \cdot \left( \frac{x \sin x}{\cos^2 x} \right) + \sec x\)
\(= x \tan x \sec x + \sec x\)
In simple words: Treat x as a constant, separate the terms with respect to variable y, and apply standard limit definitions.

Exam Tip: Always identify the active limit variable clearly when dealing with multi-variable expressions in limits.

Question 10. Evaluate: \(\lim_{x \to 0} \left( \frac{1 - \cos x \sqrt{\cos(2x)}}{x^2} \right)\)
Answer: Rationalize the numerator:
\(= \lim_{x \to 0} \left( \frac{1 - \cos x \sqrt{\cos(2x)}}{x^2} \cdot \frac{1 + \cos x \sqrt{\cos(2x)}}{1 + \cos x \sqrt{\cos(2x)}} \right)\)
\(= \lim_{x \to 0} \left( \frac{1 - \cos^2 x \cos(2x)}{x^2(1 + \cos x \sqrt{\cos(2x)})} \right)\)
Substitute \(\cos(2x) = 2\cos^2 x - 1\):
\(= \lim_{x \to 0} \left( \frac{1 - \cos^2 x(2\cos^2 x - 1)}{x^2(1 + \cos x \sqrt{\cos(2x)})} \right)\)
\(= \lim_{x \to 0} \left( \frac{1 - 2\cos^4 x + \cos^2 x}{x^2(1 + \cos x \sqrt{\cos(2x)})} \right)\)
Rearranging terms in the numerator:
\(= \lim_{x \to 0} \left( \frac{(2\cos^2 x + 1)(1 - \cos^2 x)}{x^2(1 + \cos x \sqrt{\cos(2x)})} \right)\)
Since \(1 - \cos^2 x = \sin^2 x\):
\(= \lim_{x \to 0} \left( \frac{\sin^2 x}{x^2} \right) \cdot \lim_{x \to 0} \left( \frac{2\cos^2 x + 1}{1 + \cos x \sqrt{\cos(2x)}} \right)\)
\(= (1)^2 \cdot \left( \frac{2(1)^2 + 1}{1 + (1)(1)} \right)\)
\(= 1 \cdot \frac{3}{2} = \frac{3}{2}\)
In simple words: Rationalize the numerator containing square roots, use trigonometric identities to factor out \(\sin^2 x\), and evaluate standard limits.

Exam Tip: Rationalization combined with algebraic factorization is the standard approach for limits involving radicals and trigonometric functions.