CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 03

Q.1) Evaluate \(\lim_{x \to 0} \left(\frac{9^x - 6^x - 4^x + 4}{x}\right)\)
Answer: We have \(\lim_{x \to 0} \left(\frac{9^x - 6^x - 4^x + 4}{x}\right)\)
Writing the expression with a squared denominator adjustment by multiplying and dividing, or grouping properly:
\(= \lim_{x \to 0} \left(\frac{9^x - 6^x - 4^x + 4}{x^2}\right) \times x\) - wait, let us factor the numerator:
Numerator = \(9^x - 6^x - 4^x + 4 = 3^x(3^x - 2^x) - 2^x(3^x - 2^x) = (3^x - 2^x)(3^x - 2^x) = (3^x - 2^x)^2\)
Thus, the expression becomes:
\(= \lim_{x \to 0} \left(\frac{(3^x - 2^x)^2}{x^2}\right)\)
\(= \lim_{x \to 0} \left[ \frac{3^x - 2^x}{x} \right]^2\)
\(= \lim_{x \to 0} \left[ \frac{(3^x - 1) - (2^x - 1)}{x} \right]^2\)
\(= \left( \lim_{x \to 0} \frac{3^x - 1}{x} - \lim_{x \to 0} \frac{2^x - 1}{x} \right)^2\)
Using the standard limit formula \(\lim_{x \to 0} \frac{a^x - 1}{x} = \log a\):
\(= (\log 3 - \log 2)^2\)
Using logarithmic property \(\log A - \log B = \log\left(\frac{A}{B}\right)\):
\(= \left( \log\left(\frac{3}{2}\right) \right)^2 \text{ ans.}\)
In simple words: Factor the numerator into a squared binomial, split the limit, and apply the standard exponential limit formula.

Exam Tip: Recognize symmetric groupings in exponential expressions to convert them into standard limit forms like \(\frac{a^x - 1}{x}\).

Q.2) Differentiate using first principle method \(f(x) = \cos(3x)\)
Answer: Using the first principle definition of derivative:
\(\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(= \lim_{h \to 0} \frac{\cos(3x + 3h) - \cos(3x)}{h}\)
Using the trigonometric transformation formula \(\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)\):
\(= \lim_{h \to 0} \frac{-2 \sin\left(\frac{6x + 3h}{2}\right) \sin\left(\frac{3h}{2}\right)}{h}\)
Rearranging terms to isolate the standard limit form \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \lim_{h \to 0} \left[ \frac{\sin\left(\frac{3h}{2}\right)}{\frac{3h}{2}} \times \frac{3}{2} \right] \times \left[ -2 \sin\left(\frac{6x + 3h}{2}\right) \right]\)
\(= 1 \times \left( -3 \sin(3x) \right)\)
\(f'(x) = -3 \sin(3x) \text{ ans.}\)
In simple words: Apply the difference-of-cosines formula, adjust the denominator to match the angle inside the sine function, and take the limit as \(h \to 0\).

Exam Tip: Always balance the constant multiplier (like \(\frac{3}{2}\)) when adjusting the denominator for the standard sine limit.

Q.3) Differentiate using first formula \(f(x) = \tan(2x)\)
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\tan(2x + 2h) - \tan(2x)}{h}\)
Expressing tangent in terms of sine and cosine:
\(= \lim_{h \to 0} \frac{\frac{\sin(2x + 2h)}{\cos(2x + 2h)} - \frac{\sin(2x)}{\cos(2x)}}{h}\)
Taking the common denominator and simplifying the numerator using \(\sin(A - B)\):
\(= \lim_{h \to 0} \frac{\sin(2x + 2h)\cos(2x) - \cos(2x + 2h)\sin(2x)}{h \cos(2x + 2h)\cos(2x)}\)
\(= \lim_{h \to 0} \frac{\sin(2x + 2h - 2x)}{h \cos(2x + 2h)\cos(2x)}\)
\(= \lim_{h \to 0} \left( \frac{\sin(2h)}{2h} \times \frac{2}{\cos(2x + 2h)\cos(2x)} \right)\)
Evaluating the limits as \(h \to 0\):
\(= 1 \times \frac{2}{\cos(2x)\cos(2x)}\)
\(\frac{dy}{dx} = 2 \sec^2(2x) \text{ ans.}\)
In simple words: Convert tangent to sine and cosine, combine fractions, use the sine subtraction formula, and apply the standard limit for \(\sin(2h)/(2h)\).

Exam Tip: Remember that \(\frac{1}{\cos^2 \theta} = \sec^2 \theta\), which helps quickly convert the final trigonometric expression.

Q.4) \(f(x) = \sqrt{\tan x}\), find \(f'(x)\) by first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{\tan(x + h)} - \sqrt{\tan x}}{h}\)
Rationalizing the numerator:
\(= \lim_{h \to 0} \frac{\tan(x + h) - \tan x}{h(\sqrt{\tan(x + h)} + \sqrt{\tan x})}\)
Using the identity \(\tan(x + h) - \tan x = \frac{\sin h}{\cos(x + h)\cos x}\) or rewriting as:
\(= \lim_{h \to 0} \frac{\frac{\sin h}{\cos(x+h)\cos x}}{h(\sqrt{\tan(x + h)} + \sqrt{\tan x})}\)
Separating the standard limit \(\lim_{h \to 0} \frac{\sin h}{h} = 1\):
\(= 1 \times \frac{1}{\cos^2 x \cdot 2\sqrt{\tan x}}\)
Since \(\frac{1}{\cos^2 x} = \sec^2 x\):
\(f'(x) = \frac{1}{2\sqrt{\tan x}} \sec^2 x \text{ ans.}\)
In simple words: Rationalize the square root expression, convert the difference of tangents into sine over cosines, and evaluate the standard limits.

Exam Tip: Rationalization with square roots often requires a secondary conversion of trigonometric terms to reach a clean form.

Q.5) \(f(x) = \sec^2 x\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\sec^2(x + h) - \sec^2 x}{h}\)
Expressing secant as cosine:
\(= \lim_{h \to 0} \frac{\frac{1}{\cos^2(x + h)} - \frac{1}{\cos^2 x}}{h}\)
Combining fractions:
\(= \lim_{h \to 0} \frac{\cos^2 x - \cos^2(x + h)}{h \cos^2(x + h)\cos^2 x}\)
Using the trigonometric identity \(\cos^2 A - \cos^2 B = \sin(A + B)\sin(B - A)\):
\(= \lim_{h \to 0} \frac{\sin(2x + h)\sin h}{h \cos^2(x + h)\cos^2 x}\)
Separating limits:
\(= \lim_{h \to 0} \frac{\sin h}{h} \times \lim_{h \to 0} \sin(2x + h) \times \lim_{h \to 0} \frac{1}{\cos^2(x + h)\cos^2 x}\)
\(= 1 \times \sin(2x) \times \frac{1}{\cos^4 x}\)
Since \(\sin(2x) = 2 \sin x \cos x\):
\(= \frac{2 \sin x \cos x}{\cos^4 x} = \frac{2 \sin x}{\cos^3 x} = 2 \tan x \sec^2 x \text{ ans.}\)
In simple words: Convert secant to cosine, simplify the complex fraction into a product of sines, apply the standard sine limit, and simplify to tangent and secant.

Exam Tip: Knowing the factorization \(\cos^2 A - \cos^2 B = \sin(A+B)\sin(B-A)\) saves several steps in trigonometric limit problems.

Q.6) \(f(x) = \sin(x^2)\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\sin((x + h)^2) - \sin(x^2)}{h}\)
\(= \lim_{h \to 0} \frac{\sin(x^2 + h^2 + 2hx) - \sin(x^2)}{h}\)
Using the difference formula for sine, \(\sin C - \sin D = 2 \cos\left(\frac{C + D}{2}\right) \sin\left(\frac{C - D}{2}\right)\):
\(= \lim_{h \to 0} \frac{2 \cos\left(\frac{2x^2 + h^2 + 2hx}{2}\right) \sin\left(\frac{h^2 + 2hx}{2}\right)}{h}\)
Adjusting the denominator for the sine term:
\(= \lim_{h \to 0} \left[ 2 \cos\left(\frac{2x^2 + h^2 + 2hx}{2}\right) \times \frac{\sin\left(\frac{h^2 + 2hx}{2}\right)}{\frac{h^2 + 2hx}{2}} \times \frac{h(h + 2x)}{2h} \right]\)
Evaluating the limit as \(h \to 0\):
\(= 2 \cos(x^2) \times 1 \times \frac{2x}{2}\)
\(\frac{dy}{dx} = 2x \cos(x^2) \text{ ans.}\)
In simple words: Expand the squared angle, apply the sum-to-product formula for sine, adjust the angle denominator, and evaluate the limit.

Exam Tip: This is a standard composite function derivative verified via first principles, matching the chain rule result.

Q.7) \(f(x) = \tan\sqrt{x}\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\tan(\sqrt{x + h}) - \tan(\sqrt{x})}{h}\)
Converting to sine and cosine:
\(= \lim_{h \to 0} \frac{\sin(\sqrt{x + h} - \sqrt{x})}{h \cos(\sqrt{x + h})\cos(\sqrt{x})}\)
Rationalizing the sine numerator term by multiplying and dividing by \((\sqrt{x + h} + \sqrt{x})\):
\(= \lim_{h \to 0} \left[ \frac{\sin(\sqrt{x + h} - \sqrt{x})}{\sqrt{x + h} - \sqrt{x}} \times \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{1}{\cos(\sqrt{x + h})\cos(\sqrt{x})} \right]\)
Further rationalizing \(\frac{\sqrt{x + h} - \sqrt{x}}{h}\) gives \(\frac{1}{\sqrt{x + h} + \sqrt{x}}\).
Taking the limit as \(h \to 0\):
\(= 1 \times \frac{1}{\cos(\sqrt{x})\cos(\sqrt{x})} \times \frac{1}{2\sqrt{x}}\)
\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \sec^2(\sqrt{x}) \text{ ans.}\)
In simple words: Convert to sine and cosine, rationalize the angle difference for the sine term, and simplify using standard limits.

Exam Tip: Double rationalization (first for tangents/sines, then for square roots) is required when both radicals and trigonometric functions are present.

Q.8) \(f(x) = x \cos x\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{(x + h)\cos(x + h) - x \cos x}{h}\)
Rearranging terms by grouping \(x\) terms together:
\(= \lim_{h \to 0} \frac{x[\cos(x + h) - \cos x] + h \cos(x + h)}{h}\)
\(= \lim_{h \to 0} \left[ x \cdot \frac{\cos(x + h) - \cos x}{h} + \cos(x + h) \right]\)
Using the cosine difference formula for the first term: \(\cos(x + h) - \cos x = -2 \sin\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)\)
\(= \lim_{h \to 0} \left[ x \cdot \frac{-2 \sin\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{2 \cdot \frac{h}{2}} + \cos(x + h) \right]\)
Evaluating the limit as \(h \to 0\):
\(= x (-1) \sin x + \cos x\)
\(f'(x) = -x \sin x + \cos x \text{ ans.}\)
In simple words: Group terms containing \(x\) to form the derivative of cosine, keep the remaining term, and evaluate the limits directly.

Exam Tip: This confirms the product rule for differentiation from first principles: \((uv)' = u'v + uv'\).

Q.9) \(f(x) = \frac{\sin x}{x}\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{\frac{\sin(x + h)}{x + h} - \frac{\sin x}{x}}{h}\)
Combining fractions in the numerator:
\(= \lim_{h \to 0} \frac{x \sin(x + h) - (x + h)\sin x}{h x(x + h)}\)
Expanding and rearranging:
\(= \lim_{h \to 0} \frac{x[\sin(x + h) - \sin x] - h \sin x}{h x(x + h)}\)
Splitting into two parts and applying the sine difference formula \(\sin(x + h) - \sin x = 2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)\):
\(= \lim_{h \to 0} \left[ \frac{x \cdot 2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h x(x + h)} - \frac{h \sin x}{h x(x + h)} \right]\)
Evaluating the limits as \(h \to 0\):
\(= \frac{x \cos x \cdot 1}{x^2} - \frac{\sin x}{x^2}\)
\(\frac{dy}{dx} = \frac{x \cos x - \sin x}{x^2} \text{ ans.}\)
In simple words: Take a common denominator, group terms to isolate the sine difference formula, and divide through by \(h\) to apply standard limits.

Exam Tip: This is the exact verification of the quotient rule for trigonometric functions from first principles.

Q.10) \(f(x) = \sin x - \cos x\) using first principle method.
Answer: Using the first principle definition:
\(f'(x) = \lim_{h \to 0} \frac{[\sin(x + h) - \cos(x + h)] - [\sin x - \cos x]}{h}\)
Rearranging terms by grouping sines and cosines together:
\(= \lim_{h \to 0} \frac{[\sin(x + h) - \sin x] - [\cos(x + h) - \cos x]}{h}\)
Applying sum-to-product trigonometric formulas:
\(= \lim_{h \to 0} \frac{2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right) + 2 \sin\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}\)
Factoring out common terms and adjusting the denominator:
\(= \lim_{h \to 0} \left[ \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \times \left( \cos\left(\frac{2x + h}{2}\right) + \sin\left(\frac{2x + h}{2}\right) \right) \right]\)
Taking the limit as \(h \to 0\):
\(= 1 \times (\cos x + \sin x)\)
\(f'(x) = \cos x + \sin x \text{ ans.}\)
In simple words: Separate the expression into sine and cosine differences, apply standard transformation formulas, and evaluate the limit.

Exam Tip: Derivatives of sums/differences are simply the sum/difference of their individual derivatives.