CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 01

Question 1. Given \(f(x) = \frac{x \cos x - x \sin x}{x \cos x + x \sin x}\), find \(f'(x)\).
Answer: We are given the function \(f(x) = \frac{x \cos x - x \sin x}{x \cos x + x \sin x}\).
Simplifying the function first by factoring out \(x\) in the numerator and denominator:
\(f(x) = \frac{x(\cos x - \sin x)}{x(\cos x + \sin x)} = \frac{\cos x - \sin x}{\cos x + \sin x}\)
Now, differentiating with respect to \(x\) using the quotient rule:
\(f'(x) = \frac{(\cos x + \sin x) \cdot \frac{d}{dx}(\cos x - \sin x) - (\cos x - \sin x) \cdot \frac{d}{dx}(\cos x + \sin x)}{(\cos x + \sin x)^2}\)
\(f'(x) = \frac{(\cos x + \sin x)(-\sin x - \cos x) - (\cos x - \sin x)(-\sin x + \cos x)}{(\cos x + \sin x)^2}\)
\(f'(x) = \frac{-(\cos x + \sin x)^2 - (\cos x - \sin x)(\cos x - \sin x)}{(\cos x + \sin x)^2}\)
\(f'(x) = \frac{-[(\cos x + \sin x)^2 + (\cos x - \sin x)^2]}{(\cos x + \sin x)^2}\)
Using algebraic identity \((a+b)^2 + (a-b)^2 = 2(a^2 + b^2)\):
\(f'(x) = \frac{-2(\cos^2 x + \sin^2 x)}{(\cos x + \sin x)^2}\)
Since \(\cos^2 x + \sin^2 x = 1\):
\(f'(x) = \frac{-2}{(\cos x + \sin x)^2}\)
In simple words: Simplify the fraction by canceling \(x\), then apply the quotient rule and trigonometric identities to get the derivative.

Exam Tip: Always simplify rational trigonometric functions before applying differentiation rules to save time and avoid calculation errors.

Question 2. Given \(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1\), show that \(f'(1) = 100 f'(0)\).
Answer: Given function: \(f(x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + \dots + \frac{x^2}{2} + x + 1\)
Differentiating with respect to \(x\) using the power rule \(\frac{d}{dx}(x^n) = n x^{n-1}\):
\(f'(x) = \frac{100 x^{99}}{100} + \frac{99 x^{98}}{99} + \dots + \frac{2x}{2} + 1 + 0\)
\(f'(x) = x^{99} + x^{98} + \dots + x + 1\)
Evaluating at \(x = 1\):
\(f'(1) = 1^{99} + 1^{98} + \dots + 1 + 1\)
Since there are 100 terms of 1: \(f'(1) = 100\)
Evaluating at \(x = 0\):
\(f'(0) = 0^{99} + 0^{98} + \dots + 0 + 1 = 1\)
Now, checking the condition \(f'(1) = 100 f'(0)\):
L.H.S. = \(f'(1) = 100\)
R.H.S. = \(100 \times f'(0) = 100 \times 1 = 100\)
Therefore, L.H.S. = R.H.S.
In simple words: Take the derivative term by term using the power rule, substitute \(x = 1\) and \(x = 0\), and compare both sides.

Exam Tip: Be careful with the number of terms when substituting \(x = 1\) in a polynomial series.

Question 3. If \(y = \sqrt{\frac{1 - \cos x}{1 + \cos x}}\), find \(\frac{dy}{dx}\).
Answer: Given: \(y = \sqrt{\frac{1 - \cos x}{1 + \cos x}}\)
Using half-angle formulas \(1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)\) and \(1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right)\):
\(y = \sqrt{\frac{2 \sin^2\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)}}\)
\(y = \sqrt{\tan^2\left(\frac{x}{2}\right)} = \tan\left(\frac{x}{2}\right)\)
Now, differentiating with respect to \(x\):
\(\frac{dy}{dx} = \sec^2\left(\frac{x}{2}\right) \cdot \frac{d}{dx}\left(\frac{x}{2}\right)\)
\(\frac{dy}{dx} = \sec^2\left(\frac{x}{2}\right) \times \frac{1}{2} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right)\)
In simple words: Convert trigonometric terms into half-angle tangent form using identities, simplify the square root, and then use the chain rule.

Exam Tip: Memorize standard trigonometric half-angle substitutions as they drastically simplify inverse and radical calculus problems.

Question 4. If \(f(x) = \frac{\sin(x + a)}{\cos x}\), find \(f'(x)\).
Answer: Given: \(f(x) = \frac{\sin(x + a)}{\cos x}\)
Differentiating with respect to \(x\) using the quotient rule:
\(f'(x) = \frac{\cos x \cdot \frac{d}{dx}(\sin(x + a)) - \sin(x + a) \cdot \frac{d}{dx}(\cos x)}{\cos^2 x}\)
\(f'(x) = \frac{\cos x \cdot \cos(x + a) - \sin(x + a)(-\sin x)}{\cos^2 x}\)
\(f'(x) = \frac{\cos x \cos(x + a) + \sin x \sin(x + a)}{\cos^2 x}\)
Using the cosine subtraction formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\), where \(A = x + a\) and \(B = x\):
\(f'(x) = \frac{\cos((x + a) - x)}{\cos^2 x}\)
\(f'(x) = \frac{\cos a}{\cos^2 x} = \cos a \sec^2 x\)
In simple words: Apply the quotient rule, expand using derivative formulas, and then use the cosine subtraction identity to simplify the numerator.

Exam Tip: Watch out for the negative sign when differentiating \(\cos x\), and look for compound angle formula applications in the numerator.

Question 5. Evaluate \(\lim_{x \to \infty} \left(\frac{\sqrt{3x^2 + 1} + \sqrt{2x^2 - 1}}{4x + 3}\right)\).
Answer: Given limit: \(\lim_{x \to \infty} \left(\frac{\sqrt{3x^2 + 1} + \sqrt{2x^2 - 1}}{4x + 3}\right)\)
Dividing both numerator and denominator by \(x\):
\(= \lim_{x \to \infty} \left(\frac{\frac{\sqrt{3x^2 + 1}}{x} + \frac{\sqrt{2x^2 - 1}}{x}}{\frac{4x + 3}{x}}\right)\)
Bringing \(x\) inside the square roots as \(x^2\):
\(= \lim_{x \to \infty} \left(\frac{\sqrt{\frac{3x^2 + 1}{x^2}} + \sqrt{\frac{2x^2 - 1}{x^2}}}{4 + \frac{3}{x}}\right)\)
\(= \lim_{x \to \infty} \left(\frac{\sqrt{3 + \frac{1}{x^2}} + \sqrt{2 - \frac{1}{x^2}}}{4 + \frac{3}{x}}\right)\)
Taking the limit as \(x \to \infty\) (terms with \(x\) in the denominator approach 0):
\(= \frac{\sqrt{3 + 0} + \sqrt{2 - 0}}{4 + 0} = \frac{\sqrt{3} + \sqrt{2}}{4}\)
In simple words: Divide every term by the highest power of \(x\) in the denominator, then apply the limit as \(x \to \infty\).

Exam Tip: When moving \(x\) inside a square root in the denominator, remember that \(x = \sqrt{x^2}\) for positive \(x\).

Question 6. Evaluate \(\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})\).
Answer: Given limit: \(\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})\)
Rationalizing the expression by multiplying and dividing by its conjugate:
\(= \lim_{x \to \infty} \frac{(\sqrt{x^2 + x + 1} - \sqrt{x^2 + 1})(\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1})}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}}\)
\(= \lim_{x \to \infty} \frac{(x^2 + x + 1) - (x^2 + 1)}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}}\)
\(= \lim_{x \to \infty} \frac{x}{\sqrt{x^2 + x + 1} + \sqrt{x^2 + 1}}\)
Dividing both numerator and denominator by \(x\):
\(= \lim_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} + \sqrt{1 + \frac{1}{x^2}}}\)
Taking the limit as \(x \to \infty\):
\(= \frac{1}{\sqrt{1 + 0 + 0} + \sqrt{1 + 0}} = \frac{1}{1 + 1} = \frac{1}{2}\)
In simple words: Rationalize the difference of square roots, simplify the numerator, and then divide by \(x\) to evaluate the limit.

Exam Tip: Rationalization is the standard technique to resolve infinity minus infinity indeterminate forms involving radicals.

Question 7. Evaluate \(\lim_{n \to \infty} \left(\frac{1 + 2 + 3 + \dots + n}{n^2}\right)\).
Answer: Given limit: \(\lim_{n \to \infty} \left(\frac{1 + 2 + 3 + \dots + n}{n^2}\right)\)
Using the formula for the sum of first \(n\) natural numbers \(1 + 2 + \dots + n = \frac{n(n + 1)}{2}\):
\(= \lim_{n \to \infty} \frac{n(n + 1)}{2n^2}\)
\(= \lim_{n \to \infty} \frac{n^2 + n}{2n^2}\)
Dividing numerator and denominator by \(n^2\):
\(= \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2}\)
Taking the limit as \(n \to \infty\):
\(= \frac{1 + 0}{2} = \frac{1}{2}\)
In simple words: Replace the arithmetic series sum in the numerator with its standard formula, then divide by \(n^2\) and take the limit.

Question 8. Evaluate \(\lim_{x \to -\infty} (\sqrt{x^2 - x + 1} + x)\).
Answer: Given limit: \(\lim_{x \to -\infty} (\sqrt{x^2 - x + 1} + x)\)
Let \(x = -y\). As \(x \to -\infty\), \(y \to \infty\).
Substituting \(x = -y\) into the limit:
\(= \lim_{y \to \infty} (\sqrt{(-y)^2 - (-y) + 1} + (-y))\)
\(= \lim_{y \to \infty} (\sqrt{y^2 + y + 1} - y)\)
Rationalizing by multiplying and dividing by the conjugate \((\sqrt{y^2 + y + 1} + y)\):
\(= \lim_{y \to \infty} \frac{(y^2 + y + 1) - y^2}{\sqrt{y^2 + y + 1} + y}\)
\(= \lim_{y \to \infty} \frac{y + 1}{\sqrt{y^2 + y + 1} + y}\)
Dividing numerator and denominator by \(y\):
\(= \lim_{y \to \infty} \frac{1 + \frac{1}{y}}{\sqrt{1 + \frac{1}{y} + \frac{1}{y^2}} + 1}\)
Taking the limit as \(y \to \infty\):
\(= \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}\)
In simple words: Substitute \(x = -y\) to change the limit to positive infinity, then rationalize and solve as usual.

Exam Tip: Substituting \(x = -y\) is crucial for handling limits approaching negative infinity with square roots.

Question 9. Find the derivative of \(\sqrt[3]{\sin x}\) using first principle method.
Answer: Let \(f(x) = \sqrt[3]{\sin x} = (\sin x)^{\frac{1}{3}}\)
Using the first principle formula: \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{(\sin(x + h))^{\frac{1}{3}} - (\sin x)^{\frac{1}{3}}}{h}\)
Multiplying and dividing by \(\sin(x + h) - \sin x\):
\(f'(x) = \lim_{h \to 0} \left[ \frac{(\sin(x + h))^{\frac{1}{3}} - (\sin x)^{\frac{1}{3}}}{\sin(x + h) - \sin x} \times \frac{\sin(x + h) - \sin x}{h} \right]\)
As \(h \to 0\), \(\sin(x + h) \to \sin x\). Using the formula \(\lim_{z \to a} \frac{z^n - a^n}{z - a} = n a^{n-1}\):
\(f'(x) = \frac{1}{3}(\sin x)^{\frac{1}{3} - 1} \times \lim_{h \to 0} \left( \frac{2 \sin\left(\frac{h}{2}\right) \cos\left(x + \frac{h}{2}\right)}{h} \right)\)
\(f'(x) = \frac{1}{3} (\sin x)^{-\frac{2}{3}} \times \lim_{h \to 0} \left( \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \right) \times \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right)\)
Using standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\) and evaluating the cosine limit:
\(f'(x) = \frac{1}{3} (\sin x)^{-\frac{2}{3}} \times 1 \times \cos x\)
\(f'(x) = \frac{1}{3} (\sin x)^{-\frac{2}{3}} \cos x\)
In simple words: Set up the first principle limit, use algebraic limits for fractional powers, and apply the sum-to-product sine formula along with standard trigonometric limits.

Exam Tip: Master the algebraic limit formula \(\lim_{z \to a} \frac{z^n - a^n}{z - a} = n a^{n-1}\) for first principle problems involving roots and fractional powers.

Question 10. Find the derivative of \(e^{\sqrt{\tan x}}\) using first principle method.
Answer: Let \(f(x) = e^{\sqrt{\tan x}}\)
Using the first principle formula:
\(f'(x) = \lim_{h \to 0} \frac{e^{\sqrt{\tan(x + h)}} - e^{\sqrt{\tan x}}}{h}\)
Factor out \(e^{\sqrt{\tan x}}\):
\(f'(x) = e^{\sqrt{\tan x}} \lim_{h \to 0} \frac{e^{\sqrt{\tan(x + h)} - \sqrt{\tan x}} - 1}{h}\)
Multiply and divide by \(\sqrt{\tan(x + h)} - \sqrt{\tan x}\):
\(f'(x) = e^{\sqrt{\tan x}} \lim_{h \to 0} \left[ \frac{e^{\sqrt{\tan(x + h)} - \sqrt{\tan x}} - 1}{\sqrt{\tan(x + h)} - \sqrt{\tan x}} \times \frac{\sqrt{\tan(x + h)} - \sqrt{\tan x}}{h} \right]\)
As \(h \to 0\), \(\sqrt{\tan(x + h)} \to \sqrt{\tan x}\). Using \(\lim_{z \to 0} \frac{e^z - 1}{z} = 1\):
\(f'(x) = e^{\sqrt{\tan x}} \times 1 \times \lim_{h \to 0} \left[ \frac{\sqrt{\tan(x + h)} - \sqrt{\tan x}}{h} \times \frac{\sqrt{\tan(x + h)} + \sqrt{\tan x}}{\sqrt{\tan(x + h)} + \sqrt{\tan x}} \right]\)
\(f'(x) = e^{\sqrt{\tan x}} \lim_{h \to 0} \left[ \frac{\tan(x + h) - \tan x}{h(\sqrt{\tan(x + h)} + \sqrt{\tan x})} \right]\)
Using identity \(\tan(x + h) - \tan x = \frac{\sin h}{\cos(x + h) \cos x}\), or directly expanding: Since \(\lim_{h \to 0} \frac{\tan h}{h} = 1\) and using compound angle expansion:
\(f'(x) = e^{\sqrt{\tan x}} \times \frac{\sec^2 x}{2\sqrt{\tan x}}\)
\(f'(x) = \frac{e^{\sqrt{\tan x}} \sec^2 x}{2\sqrt{\tan x}}\)
In simple words: Use the exponential limit definition combined with rationalization and trigonometric limits to evaluate the derivative step by step from first principles.

Exam Tip: For exponential functions in first principles, use the standard limit \(\lim_{z \to 0} \frac{e^z - 1}{z} = 1\) after factoring out the base function term.