CBSE Class 11 Mathematics Revision Assignment Set 03

Question 1. Prove by the principle of Mathematical induction for all natural numbers: \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}\)
Answer: Let the given statement be \(P(n)\):
\(P(n) : \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}\)
Step 1: For \(n = 1\),
LHS = \(\frac{1}{1 \cdot 2} = \frac{1}{2}\)
RHS = \(\frac{1}{1+1} = \frac{1}{2}\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume that \(P(k)\) is true for some positive integer \(k\), i.e.,
\(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} = \frac{k}{k+1}\)
Step 3: We need to prove that \(P(k+1)\) is true, i.e.,
\(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2}\)
Starting with the LHS of \(P(k+1)\):
\(= \left[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{k(k+1)} \right] + \frac{1}{(k+1)(k+2)}\)
\(= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}\)
\(= \frac{1}{k+1} \left( k + \frac{1}{k+2} \right)\)
\(= \frac{1}{k+1} \left( \frac{k(k+2) + 1}{k+2} \right)\)
\(= \frac{1}{k+1} \left( \frac{k^2 + 2k + 1}{k+2} \right)\)
\(= \frac{1}{k+1} \frac{(k+1)^2}{k+2}\)
\(= \frac{k+1}{k+2}\)
Which is equal to the RHS of \(P(k+1)\). Thus, \(P(k+1)\) is true whenever \(P(k)\) is true. Hence, by the principle of mathematical induction, \(P(n)\) is true for all natural numbers \(n\).
In simple words: We check the formula for 1, assume it works for \(k\), and prove it works for \(k+1\) by algebraic substitution.

Exam Tip: Always clearly state the three steps: base case (\(n=1\)), induction hypothesis (\(n=k\)), and induction step (\(n=k+1\)).

Question 2. Prove by the principle of Mathematical induction for all natural numbers: \(\sin\theta + \sin 2\theta + \sin 3\theta + \dots + \sin n\theta = \frac{\sin\left(\frac{n\theta}{2}\right) \sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Answer: Let \(P(n)\) be the statement:
\(P(n) : \sin\theta + \sin 2\theta + \dots + \sin n\theta = \frac{\sin\left(\frac{n\theta}{2}\right) \sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Step 1: For \(n = 1\),
LHS = \(\sin\theta\)
RHS = \(\frac{\sin\left(\frac{\theta}{2}\right) \sin\left(\frac{2\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} = \sin\theta\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true:
\(\sin\theta + \sin 2\theta + \dots + \sin k\theta = \frac{\sin\left(\frac{k\theta}{2}\right) \sin\left(\frac{(k+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Step 3: To prove for \(n = k+1\):
LHS = \([\sin\theta + \dots + \sin k\theta] + \sin(k+1)\theta\)
\(= \frac{\sin\left(\frac{k\theta}{2}\right) \sin\left(\frac{(k+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} + \sin(k+1)\theta\)
\(= \frac{1}{\sin\left(\frac{\theta}{2}\right)} \left[ \sin\left(\frac{k\theta}{2}\right)\sin\left(\frac{(k+1)\theta}{2}\right) + \sin\left(\frac{\theta}{2}\right)\sin((k+1)\theta) \right]\)
Using the product-to-sum formula \(2\sin A \sin B = \cos(A-B) - \cos(A+B)\), we simplify the bracket to obtain:
\(= \frac{\sin\left(\frac{(k+1)\theta}{2}\right) \sin\left(\frac{(k+2)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}\)
Thus, \(P(k+1)\) is true. Hence proved by induction.
In simple words: Trigonometric induction follows the exact same three steps as algebraic induction, utilizing standard trigonometric identities.

Exam Tip: Be careful with angle fractions and trigonometric transformation identities during the induction step.

Question 3. Prove by the principle of Mathematical induction for all natural numbers: \((11^{n+2} + 12^{2n+1})\) is divisible by 133.
Answer: Let \(P(n) : 11^{n+2} + 12^{2n+1} = 133k\) for some integer \(k\).
Step 1: For \(n = 1\),
\(11^{1+2} + 12^{2(1)+1} = 11^3 + 12^3 = 1331 + 1728 = 3059\)
Since \(\frac{3059}{133} = 23\), it is divisible by 133. Thus, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true, i.e., \(11^{k+2} + 12^{2k+1} = 133m\) for some integer \(m\).
\(11^{k+2} = 133m - 12^{2k+1}\)
Step 3: For \(n = k+1\),
\(11^{(k+1)+2} + 12^{2(k+1)+1} = 11 \cdot 11^{k+2} + 12^2 \cdot 12^{2k+1}\)
Substitute \(11^{k+2} = 133m - 12^{2k+1}\):
\(= 11(133m - 12^{2k+1}) + 144 \cdot 12^{2k+1}\)
\(= 11 \times 133m - 11 \cdot 12^{2k+1} + 144 \cdot 12^{2k+1}\)
\(= 11 \times 133m + (144 - 11) \cdot 12^{2k+1}\)
\(= 11 \times 133m + 133 \cdot 12^{2k+1}\)
\(= 133(11m + 12^{2k+1})\)
This expression is clearly a multiple of 133. Hence, \(P(k+1)\) is true, and the statement holds for all natural numbers \(n\).
In simple words: Express one term in terms of the other using the induction assumption so that 133 factors out cleanly.

Exam Tip: For divisibility proofs, always isolate one exponent term from the inductive hypothesis and substitute it into the \(k+1\) expression.

Question 4. Find the conjugate of \(\frac{(3+i\sqrt{2})(3-i\sqrt{2})}{( \sqrt{3}+i\sqrt{2})-(\sqrt{3}-i\sqrt{2})}\)
Answer: Let \(z = \frac{(3+i\sqrt{2})(3-i\sqrt{2})}{(\sqrt{3}+i\sqrt{2})-(\sqrt{3}-i\sqrt{2})}\)
Numerator = \((3)^2 - (i\sqrt{2})^2 = 9 - (-2) = 9 + 2 = 11\)
Denominator = \(\sqrt{3} + i\sqrt{2} - \sqrt{3} + i\sqrt{2} = 2i\sqrt{2}\)
So, \(z = \frac{11}{2i\sqrt{2}}\)
Rationalizing the denominator:
\(z = \frac{11}{2\sqrt{2}i} \times \frac{-i}{-i} = \frac{-11i}{2\sqrt{2}(-i^2)} = \frac{-11i}{2\sqrt{2}} = -\frac{11\sqrt{2}}{4}i\)
The conjugate \(\bar{z}\) is:
\(\bar{z} = \frac{11\sqrt{2}}{4}i\)
In simple words: Simplify the numerator and denominator separately using algebra, then multiply numerator and denominator by \(i\) to remove \(i\) from the bottom.

Exam Tip: Remember that the conjugate of \(x + iy\) is \(x - iy\). Change the sign of the imaginary part only.

Question 5. If \(z = -5 + 2\sqrt{-4}\) then show that \(z^2 + 10z + 41 = 0\) and hence evaluate \((z^4 + 9z^3 + 35z^2 - z + 4)\)
Answer: Given \(z = -5 + 2\sqrt{-4} = -5 + 2(2i) = -5 + 4i\)
\(z + 5 = 4i\)
Squaring both sides:
\((z + 5)^2 = (4i)^2\)
\(z^2 + 10z + 25 = -16\)
\(z^2 + 10z + 41 = 0\)
Now, to evaluate \(z^4 + 9z^3 + 35z^2 - z + 4\), we perform polynomial long division by \(z^2 + 10z + 41\):
\(z^4 + 9z^3 + 35z^2 - z + 4 = (z^2 + 10z + 41)(z^2 - z + 4) + 40\)
Since \(z^2 + 10z + 41 = 0\):
\(z^4 + 9z^3 + 35z^2 - z + 4 = 0 \cdot (z^2 - z + 4) + 40 = 40\)
In simple words: Rearrange the given complex number equation to form a quadratic equation equal to zero, then divide the target polynomial by this quadratic to find the numerical remainder.

Exam Tip: Whenever asked to evaluate a high-degree polynomial given a quadratic relation, use division algorithm or synthetic division to substitute the quadratic expression as zero.

Question 6. Express the following complex number in polar form: (i) \((\sqrt{3} + i)\) (ii) \(\sin 60^\circ + i \cos 60^\circ\) (iii) \((1 - \cos\theta + i\sin\theta\) (iv) \(\frac{1+2i}{1-3i}\) (v) \((1+i\sqrt{3})\)
Answer:
(i) Let \(z = \sqrt{3} + i\). Here \(r = \sqrt{(\sqrt{3})^2 + 1^2} = 2\).
\(\cos\theta = \frac{\sqrt{3}}{2}\), \(\sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}\).
Polar form: \(2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\)

(ii) Let \(z = \sin 60^\circ + i\cos 60^\circ = \cos 30^\circ + i\sin 30^\circ\).
Here \(r = 1\), argument \(\theta = \frac{\pi}{6}\).
Polar form: \(1\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)\)

(iii) Let \(z = 1 - \cos\theta + i\sin\theta = 2\sin^2\frac{\theta}{2} + i\left(2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right)\).
Magnitude \(r = 2\sin\frac{\theta}{2}\), argument \(\theta' = \frac{\pi}{2} - \frac{\theta}{2}\).
Polar form: \(2\sin\frac{\theta}{2} \left[ \cos\left(\frac{\pi-\theta}{2}\right) + i\sin\left(\frac{\pi-\theta}{2}\right) \right]\)

(iv) Let \(z = \frac{1+2i}{1-3i} = \frac{(1+2i)(1+3i)}{(1-3i)(1+3i)} = \frac{1 + 3i + 2i - 6}{1 + 9} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i\).
\(r = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \frac{1}{\sqrt{2}}\).
Since it lies in the second quadrant, \(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).
Polar form: \(\frac{1}{\sqrt{2}}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)\)

(v) Let \(z = 1 + i\sqrt{3}\). Here \(r = 2\), \(\cos\theta = \frac{1}{2}\), \(\sin\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}\).
Polar form: \(2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\)
In simple words: Find the magnitude \(r = \sqrt{x^2+y^2}\) and the principal argument \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\), then write as \(r(\cos\theta + i\sin\theta)\).

Exam Tip: Always verify the correct quadrant of the complex number before determining its argument.

Question 7. Solve for \(x\): \(21x^2 - 28x + 10 = 0\)
Answer: Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 21\), \(b = -28\), \(c = 10\):
\(Discriminant (\Delta) = (-28)^2 - 4(21)(10) = 784 - 840 = -56\)
\(x = \frac{-(-28) \pm \sqrt{-56}}{2(21)} = \frac{28 \pm \sqrt{56}i}{42} = \frac{28 \pm 2\sqrt{14}i}{42}\)
\(x = \frac{14 \pm \sqrt{14}i}{21} = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i\)
In simple words: Apply the standard quadratic formula when factoring is not straightforward, handling the negative discriminant with the imaginary unit \(i\).

Exam Tip: When the discriminant is negative, introduce \(i\) by writing \(\sqrt{-\Delta} = \sqrt{\Delta}i\).

Question 8. Solve for \(x\): \(x^2 + 7ix + 6 = 0\)
Answer: Given \(x^2 + 7ix + 6 = 0\)
Substitute \(6 = -6i^2\):
\(x^2 + 7ix - 6i^2 = 0\)
Factoring the quadratic expression:
\(x^2 + 6ix + ix - 6i^2 = 0\)
\(x(x + 6i) + i(x + 6i) = 0\)
\((x + i)(x + 6i) = 0\)
Therefore, \(x = -i\) or \(x = -6i\).
In simple words: Rewrite the constant term using \(i^2 = -1\) so the quadratic expression can be factored easily.

Exam Tip: Replacing constants with negative multiples of \(i^2\) is a powerful technique for factoring equations with complex coefficients.

Question 9. Solve for \(x\): \(2x^2 - (3+7i)x - (3-9i) = 0\)
Answer: Here \(a = 2\), \(b = -(3+7i)\), \(c = -(3-9i)\).
Discriminant \(\Delta = b^2 - 4ac = [-(3+7i)]^2 - 4(2)[-(3-9i)]\)
\(\Delta = (9 - 49 + 42i) + 8(3 - 9i) = (-40 + 42i) + (24 - 72i) = -16 - 30i\)
Let \(\sqrt{\Delta} = x + iy \implies (x+iy)^2 = -16 - 30i\)
Solving for \(x\) and \(y\) gives \(x = \pm 3\), \(y = \mp 5\) (taking appropriate signs).
Using the quadratic formula: \(x = \frac{(3+7i) \pm (3 - 5i)}{4}\)
Case 1: \(x = \frac{3 + 7i + 3 - 5i}{4} = \frac{6 + 2i}{4} = \frac{3}{2} + \frac{1}{2}i\)
Case 2: \(x = \frac{3 + 7i - 3 + 5i}{4} = \frac{12i}{4} = 3i\)
Solutions are \(x = \frac{3}{2} + \frac{1}{2}i\) and \(x = 3i\).
In simple words: Compute the discriminant by expanding complex squares, find its square root, and apply the quadratic formula.

Exam Tip: Finding the square root of a complex discriminant requires equating real and imaginary parts.

Question 10. Solve: \(\frac{3n+1}{3} - \frac{4x+5}{6} \leq \frac{4x=1}{6} - \frac{2x+3}{2}\) (Note: \(4x=1\) is a typo in source for \(4x-1\))
Answer: Corrected inequality: \(\frac{3n+1}{3} - \frac{4x+5}{6} \leq \frac{4x-1}{6} - \frac{2x+3}{2}\)
Multiply the entire inequality by 6 (LCM of 3, 6, 2):
\(2(3n+1) - (4x+5) \leq (4x-1) - 3(2x+3)\)
\(6n + 2 - 4x - 5 \leq 4x - 1 - 6x - 9\)
\(6n - 4x - 3 \leq -2x - 10\)
Assuming variable is \(x\): \(6x - 4x - 3 \leq -2x - 10 \implies 2x - 3 \leq -2x - 10\) (or treating \(n\) as constant parameter if interpreted as linear inequality in \(x\)). Standardizing for variable \(x\):
\(4x \leq -7 \implies x \leq -\frac{7}{4}\)
In simple words: Clear fractions by multiplying by the LCM, simplify both sides, and isolate the variable \(x\).

Question 11. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 liters of the 8% solution, how many liters of the 2% solution will have to be added?
Answer: Let \(x\) liters of 2% boric acid solution be added.
Total volume of mixture = \((640 + x)\) liters.
Amount of boric acid in the mixture = \(0.02x + 0.08(640) = 0.02x + 51.2\)
The percentage of boric acid in the mixture must be between 4% and 6%:
\(4 < \frac{0.02x + 51.2}{640 + x} \times 100 < 6\)
\(0.04(640 + x) < 0.02x + 51.2 < 0.06(640 + x)\)
Solving the left inequality:
\(25.6 + 0.04x < 0.02x + 51.2\)
\(0.02x < 25.6 \implies x < 1280\)
Solving the right inequality:
\(0.02x + 51.2 < 38.4 + 0.06x\)
\(12.8 < 0.04x \implies x > 320\)
Thus, the volume of the 2% solution to be added is strictly between 320 liters and 1280 liters (\(320 < x < 1280\)).
In simple words: Set up a compound inequality representing the total boric acid divided by total volume, bounded between 4% and 6%.
Exam Tip: For mixture problems, equate total pure substance (acid/solute) before and after mixing.

Question 12. Find the region of the graph where all the inequalities \((x + 2y) \geq 0\), \((2x + y) \leq 4\), \(x \geq 0\) and \(y \leq 2\) hold good. Also find the ordered pairs of the vertices of the region.
Answer: 1. Graph the boundary lines:
- \(x + 2y = 0\) passes through \((0,0)\) and \((2,-1)\).
- \(2x + y = 4\) passes through \((2,0)\) and \((0,4)\).
- \(x = 0\) (Y-axis), \(y = 2\) (horizontal line).
2. Determine the feasible region by testing test points.
3. Find the vertices by solving the intersection of boundary lines:
- Intersection of \(x = 0\) and \(y = 2\) is \((0,2)\).
- Intersection of \(2x + y = 4\) and \(y = 2\) gives \(2x + 2 = 4 \implies x = 1\), vertex \((1,2)\).
- Intersection of \(2x + y = 4\) and \(x + 2y = 0\) gives \(x = -4, y = 12\) (outside feasible region).
- Intersection of \(x = 0\) and \(x + 2y = 0\) is \((0,0)\).
- Intersection of \(2x + y = 4\) and \(x = 0\) is \((0,4)\) (not satisfying \(y \leq 2\)).
Vertices of the feasible region are \((0,0)\), \((0,2)\), \((1,2)\), and \((2,0)\).
In simple words: Shade the region that satisfies all given inequality conditions simultaneously and find the corner points.

Exam Tip: Always test the origin \((0,0)\) wherever possible to determine which side of the line to shade (provided the line does not pass through the origin).

Question 13. Solve the system of linear inequations graphically: \(x + 2y \leq 8\), \(2x + y \geq 2\), \(x - y \leq 1\), \(x \geq 0\), \(y \geq 0\)
Answer: 1. Draw the boundary lines corresponding to the inequalities:
- \(x + 2y = 8\) (passes through \((0,4)\) and \((8,0)\), shade towards origin).
- \(2x + y = 2\) (passes through \((0,2)\) and \((1,0)\), shade away from origin).
- \(x - y = 1\) (passes through \((1,0)\) and \((0,-1)\), shade towards origin).
- \(x \geq 0, y \geq 0\) restrict the solution to the first quadrant.
2. The intersection of these shaded half-planes forms a closed polygon in the first quadrant.
Vertices are obtained by pairwise intersection of the boundary lines: \((1,0), (4,0), (2,3), (0,2),\) and \((0,4)\).
In simple words: Graph each line, shade the valid side, and find the overlapping polygon region.

Exam Tip: Label all intersection points clearly on your graph as they define the boundaries of the solution set.

Question 14. Evaluate: (i) \(\lim_{x \to \pi/4} \frac{\csc^2 x - 2}{\cot x - 1}\) (ii) \(\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x}\) (iii) \(\lim_{x \to 0} \frac{\csc x - \cot x}{x}\) (iv) \(\lim_{x \to \pi/2} \left(\frac{\pi}{2} - x\right)\tan x\)
Answer:
(i) \(\lim_{x \to \pi/4} \frac{\csc^2 x - 2}{\cot x - 1} = \lim_{x \to \pi/4} \frac{(1 + \cot^2 x) - 2}{\cot x - 1} = \lim_{x \to \pi/4} \frac{\cot^2 x - 1}{\cot x - 1}\)
\(= \lim_{x \to \pi/4} (\cot x + 1) = \cot\frac{\pi}{4} + 1 = 1 + 1 = 2\)

(ii) \(\lim_{x \to 0} \frac{1 - \cos 5x}{1 - \cos 6x} = \lim_{x \to 0} \frac{2\sin^2\left(\frac{5x}{2}\right)}{2\sin^2\left(\frac{6x}{2}\right)} = \lim_{x \to 0} \left( \frac{\sin\left(\frac{5x}{2}\right)}{\frac{5x}{2}} \times \frac{5}{2} \div \left(\frac{\sin 3x}{3x} \times 3\right) \right)^2\)
\(= \left( \frac{5/2}{3} \right)^2 = \left( \frac{5}{6} \right)^2 = \frac{25}{36}\)

(iii) \(\lim_{x \to 0} \frac{\csc x - \cot x}{x} = \lim_{x \to 0} \frac{\frac{1}{\sin x} - \frac{\cos x}{\sin x}}{x} = \lim_{x \to 0} \frac{1 - \cos x}{x \sin x}\)
\(= \lim_{x \to 0} \frac{2\sin^2(x/2)}{x \cdot 2\sin(x/2)\cos(x/2)} = \lim_{x \to 0} \frac{\tan(x/2)}{x} = \lim_{x \to 0} \frac{1}{2} \cdot \frac{\tan(x/2)}{x/2} = \frac{1}{2} \times 1 = \frac{1}{2}\)

(iv) Let \(\frac{\pi}{2} - x = \theta\). As \(x \to \frac{\pi}{2}\), \(\theta \to 0\). Also \(x = \frac{\pi}{2} - \theta\).
\(\lim_{\theta \to 0} \theta \tan\left(\frac{\pi}{2} - \theta\right) = \lim_{\theta \to 0} \theta \cot\theta = \lim_{\theta \to 0} \frac{\theta}{\tan\theta} = 1\)
In simple words: Use trigonometric identities and standard limit formulas like \(\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1\) to resolve indeterminate forms.

Exam Tip: Convert secant, cosecant, and cotangent functions into sine and cosine when stuck in trigonometric limits.

Question 15. Evaluate: (i) \(\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}\) (ii) \(\lim_{x \to 0} \frac{x}{\sqrt{1+x} - 1}\) (iii) \(\lim_{x \to 4} \frac{3 - \sqrt{5+x}}{1 - \sqrt{5-x}}\) (iv) \(\lim_{x \to a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}\) (v) \(\lim_{x \to a} \frac{x\sqrt{x} - a\sqrt{a}}{x - a}\) (vi) \(\lim_{x \to 0} \frac{(1+x)^6 - 1}{(1+x)^2 - 1}\)
Answer:
(i) \(\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} = \lim_{x \to 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}\)
\(= \lim_{x \to 0} \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \lim_{x \to 0} \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})} = \frac{2}{1+1} = 1\)

(ii) \(\lim_{x \to 0} \frac{x(\sqrt{1+x} + 1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)} = \lim_{x \to 0} \frac{x(\sqrt{1+x}+1)}{1+x-1} = \lim_{x \to 0} (\sqrt{1+x} + 1) = 2\)

(iii) Rationalizing both numerator and denominator:
\(\lim_{x \to 4} \frac{(9 - (5+x))(1 + \sqrt{5-x})}{(1 - (5-x))(3 + \sqrt{5+x})} = \lim_{x \to 4} \frac{(4-x)(1 + \sqrt{5-x})}{(x-4)(3 + \sqrt{5+x})} = \lim_{x \to 4} \frac{-(1+\sqrt{5-x})}{3+\sqrt{5+x}} = \frac{-(1+1)}{3+3} = -\frac{2}{6} = -\frac{1}{3}\)

(iv) Rationalize numerator and denominator separately, yielding limit = \(\frac{2\sqrt{3}}{3}\).

(v) Using formula \(\lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1}\):
\(\lim_{x \to a} \frac{x^{3/2} - a^{3/2}}{x - a} = \frac{3}{2}a^{(3/2)-1} = \frac{3}{2}\sqrt{a}\)

(vi) \(\lim_{x \to 0} \left[ \frac{(1+x)^6 - 1}{(1+x) - 1} \div \frac{(1+x)^2 - 1}{(1+x) - 1} \right] = \frac{6(1)^5}{2(1)^1} = \frac{6}{2} = 3\)
In simple words: Rationalize radical expressions or apply standard algebraic limit formulas to evaluate limits.

Exam Tip: Memorize the standard algebraic limit formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) as it saves significant time.