CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 07

TYPE: 2 FACTORIZE

Formula: \(a^2 - b^2, a^3 - b^3, a^4 - b^4\), quadratic equation, cubic (hit & trial) L.C.M.

Q.1) Evaluate: \(\lim_{x \to 3} \left[ \frac{x^3 - 7x^2 + 15x - 9}{x^4 - 5x^3 + 27x - 27} \right]\)
Answer: Since \(x = 3\) makes both the numerator and denominator zero, \((x - 3)\) is a factor of both polynomials.
Dividing the numerator \(x^3 - 7x^2 + 15x - 9\) by \((x - 3)\), we get \((x - 3)(x^2 - 4x + 3)\).
Dividing the denominator \(x^4 - 5x^3 + 27x - 27\) by \((x - 3)\), we get \((x - 3)(x^3 - 2x^2 - 6x + 9)\).
\(= \lim_{x \to 3} \left[ \frac{(x - 3)(x^2 - 4x + 3)}{(x - 3)(x^3 - 2x^2 - 6x + 9)} \right]\)
Cancel \((x - 3)\):
\(= \lim_{x \to 3} \left[ \frac{x^2 - 4x + 3}{x^3 - 2x^2 - 6x + 9} \right]\)
Further factorizing the numerator and denominator (since \((x - 3)\) is again a factor):
Numerator = \((x - 1)(x - 3)\)
Denominator = \((x - 3)(x^2 + x - 3)\)
\(= \lim_{x \to 3} \left[ \frac{(x - 1)(x - 3)}{(x - 3)(x^2 + x - 3)} \right]\)
\(= \lim_{x \to 3} \left[ \frac{x - 1}{x^2 + x - 3} \right]\)
Substitute \(x = 3\):
\(= \frac{3 - 1}{3^2 + 3 - 3} = \frac{2}{9} \text{ ans.}\)
In simple words: Factor out \((x - 3)\) from both the numerator and denominator, simplify the rational expression, and then substitute the limit value.

Exam Tip: If direct substitution yields the indeterminate form \(\frac{0}{0}\), look for common polynomial factors using synthetic division or polynomial long division.

Q.2) Evaluate: \(\lim_{x \to \sqrt{2}} \left[ \frac{x^4 - 4}{x^2 + 3\sqrt{2}x - 8} \right]\)
Answer: \(= \lim_{x \to \sqrt{2}} \left[ \frac{(x^2)^2 - (2)^2}{x^2 + 4\sqrt{2}x - \sqrt{2}x - 8} \right]\)
\(= \lim_{x \to \sqrt{2}} \left[ \frac{(x^2 + 2)(x^2 - 2)}{x(x + 4\sqrt{2}) - \sqrt{2}(x + 4\sqrt{2})} \right]\)
\(= \lim_{x \to \sqrt{2}} \left[ \frac{(x^2 + 2)(x + \sqrt{2})(x - \sqrt{2})}{(x + 4\sqrt{2})(x - \sqrt{2})} \right]\)
Cancel \((x - \sqrt{2})\) from numerator and denominator:
\(= \frac{((\sqrt{2})^2 + 2)(\sqrt{2} + \sqrt{2})}{\sqrt{2} + 4\sqrt{2}}\)
\(= \frac{(4)(2\sqrt{2})}{5\sqrt{2}}\)
\(= \frac{8}{5} \text{ ans.}\)
In simple words: Factor the difference of squares in the numerator and split the middle term in the denominator to cancel out the zero-making factor.

Exam Tip: Pay close attention to radical terms when factoring quadratic denominators with irrational coefficients.

TYPE: 3 RATIONALIZE

When rationalize: \((\sqrt{a} - \sqrt{b})\); \((\sqrt{a} - \text{function})\); \((\text{function} - \sqrt{b})\)

Q.3) Evaluate: \(\lim_{x \to 4} \left[ \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}} \right]\)
Answer: Rationalize both numerator and denominator simultaneously:
\(= \lim_{x \to 4} \left[ \frac{(3 - \sqrt{5 + x})(3 + \sqrt{5 + x})(1 + \sqrt{5 - x})}{(1 - \sqrt{5 - x})(1 + \sqrt{5 - x})(3 + \sqrt{5 + x})} \right]\)
\(= \lim_{x \to 4} \left[ \frac{(9 - (5 + x))(1 + \sqrt{5 - x})}{(1 - (5 - x))(3 + \sqrt{5 + x})} \right]\)
\(= \lim_{x \to 4} \left[ \frac{(4 - x)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})} \right]\)
\(= \lim_{x \to 4} \left[ \frac{-(x - 4)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})} \right]\)
Cancel \((x - 4)\):
\(= \frac{-(1 + \sqrt{5 - 4})}{3 + \sqrt{5 + 4}}\)
\(= \frac{-(1 + 1)}{3 + 3} = \frac{-2}{6} = -\frac{1}{3} \text{ ans.}\)
In simple words: Multiply both numerator and denominator by the conjugate of the numerator and the conjugate of the denominator to eliminate the square roots.

Exam Tip: Remember to factor out a negative sign when reversing terms like \((4 - x)\) to \(-(x - 4)\).

Q.4) Evaluate: \(\lim_{x \to a} \left[ \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} \right]\)
Answer: Rationalize both numerator and denominator:
\(= \lim_{x \to a} \left[ \frac{(\sqrt{a + 2x} - \sqrt{3x})(\sqrt{a + 2x} + \sqrt{3x})(\sqrt{3a + x} + 2\sqrt{x})}{(\sqrt{3a + x} - 2\sqrt{x})(\sqrt{3a + x} + 2\sqrt{x})(\sqrt{a + 2x} + \sqrt{3x})} \right]\)
\(= \lim_{x \to a} \left[ \frac{(a + 2x - 3x)(\sqrt{3a + x} + 2\sqrt{x})}{(3a + x - 4x)(\sqrt{a + 2x} + \sqrt{3x})} \right]\)
\(= \lim_{x \to a} \left[ \frac{(a - x)(\sqrt{3a + x} + 2\sqrt{x})}{(3a - 3x)(\sqrt{a + 2x} + \sqrt{3x})} \right]\)
\(= \lim_{x \to a} \left[ \frac{(a - x)(\sqrt{3a + x} + 2\sqrt{x})}{3(a - x)(\sqrt{a + 2x} + \sqrt{3x})} \right]\)
Cancel \((a - x)\):
\(= \frac{\sqrt{3a + a} + 2\sqrt{a}}{3(\sqrt{a + 2a} + \sqrt{3a})}\)
\(= \frac{\sqrt{4a} + 2\sqrt{a}}{3(\sqrt{3a} + \sqrt{3a})}\)
\(= \frac{2\sqrt{a} + 2\sqrt{a}}{3(2\sqrt{3a})}\)
\(= \frac{4\sqrt{a}}{6\sqrt{3a}} = \frac{2}{3\sqrt{3}} \text{ ans.}\)
In simple words: Multiply by the conjugates of both the numerator and denominator simultaneously to simplify algebraic expressions containing square roots.

Exam Tip: Group like radical terms carefully before performing final algebraic cancellations.

Q.5) Evaluate: \(\lim_{x \to 1} \left[ \frac{(2x - 3)(\sqrt{x} - 1)}{2x^2 + x - 3} \right]\)
Answer: Factorize the denominator \(2x^2 + x - 3 = (2x + 3)(x - 1)\):
\(= \lim_{x \to 1} \left[ \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(x - 1)} \right]\)
Write \((x - 1)\) as \((\sqrt{x} - 1)(\sqrt{x} + 1)\):
\(= \lim_{x \to 1} \left[ \frac{(2x - 3)(\sqrt{x} - 1)}{(2x + 3)(\sqrt{x} - 1)(\sqrt{x} + 1)} \right]\)
Cancel \((\sqrt{x} - 1)\):
\(= \lim_{x \to 1} \left[ \frac{2x - 3}{(2x + 3)(\sqrt{x} + 1)} \right]\)
Substitute \(x = 1\):
\(= \frac{2(1) - 3}{(2(1) + 3)(\sqrt{1} + 1)}\)
\(= \frac{2 - 3}{(5)(1 + 1)} = \frac{-1}{(5)(2)} = -\frac{1}{10} \text{ ans.}\)
In simple words: Express the linear term in the denominator as a difference of squares involving square roots so it can cancel the numerator term.

Exam Tip: Recognizing that \(x - 1 = (\sqrt{x})^2 - (1)^2\) is a key technique in limits involving square roots.

TYPE: 4 \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}\)

Q.6) Evaluate: \(\lim_{x \to 2} \left[ \frac{x^{10} - 1024}{x^5 - 32} \right]\)
Answer: \(= \lim_{x \to 2} \left[ \frac{x^{10} - 2^{10}}{x^5 - 2^5} \right]\)
Divide numerator and denominator by \((x - 2)\):
\(= \frac{\lim_{x \to 2} \left( \frac{x^{10} - 2^{10}}{x - 2} \right)}{\lim_{x \to 2} \left( \frac{x^5 - 2^5}{x - 2} \right)}\)
Using the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}\):
\(= \frac{10(2)^{10-1}}{5(2)^{5-1}}\)
\(= \frac{10(2)^9}{5(2)^4}\)
\(= 2 \cdot (2)^{9-4} = 2 \cdot 2^5 = 2 \cdot 32 = 64 \text{ ans.}\)
In simple words: Rewrite constants as powers of 2, divide by \((x - 2)\), and apply the standard limit formula for power functions.

Exam Tip: Memorize the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1}\) to solve algebraic limit problems quickly.

Q.7) Evaluate: \(\lim_{x \to a} \left[ \frac{(x + 2)^{5/3} - (a + 2)^{5/3}}{x - a} \right]\)
Answer: Substitute \(x + 2 = y\). When \(x \to a\), then \(y \to a + 2\). Also, \(x - a = y - (a + 2\)).
\(= \lim_{y \to a+2} \left[ \frac{y^{5/3} - (a + 2)^{5/3}}{y - (a + 2)} \right]\)
Applying the standard formula \(\lim_{y \to k} \frac{y^n - k^n}{y - k} = nk^{n-1}\):
\(= \frac{5}{3}(a + 2)^{\frac{5}{3} - 1}\)
\(= \frac{5}{3}(a + 2)^{2/3} \text{ ans.}\)
In simple words: Use substitution to transform the expression into the standard formula format, then apply the power rule for limits.

Exam Tip: Substitution is very effective when binomial terms like \((x + k)\) appear inside fractional powers.

Q.8) Evaluate: \(\lim_{x \to 0} \left[ \frac{(1 + x)^6 - 1}{(1 + x)^2 - 1} \right]\)
Answer: Put \(1 + x = y\). When \(x \to 0\), then \(y \to 1\).
\(= \lim_{y \to 1} \left[ \frac{y^6 - 1}{y^2 - 1} \right]\)
Divide numerator and denominator by \((y - 1)\):
\(= \frac{\lim_{y \to 1} \left( \frac{y^6 - 1}{y - 1} \right)}{\lim_{y \to 1} \left( \frac{y^2 - 1}{y - 1} \right)}\)
Applying the standard formula:
\(= \frac{6(1)^{6-1}}{2(1)^{2-1}}\)
\(= \frac{6}{2} = 3 \text{ ans.}\)
In simple words: Substitute \(1 + x = y\) to simplify the fraction, then divide numerator and denominator by \((y - 1)\) to use the standard limit formula.

Exam Tip: Always check if changing the variable simplifies a complex expression into a standard textbook form.

Q.9) Evaluate: \(\lim_{x \to 1} \left[ \frac{(x + x^2 + x^3 + \dots + x^n) - n}{x - 1} \right]\)
Answer: Rewrite \(-n\) as \(-(1 + 1 + 1 + \dots n \text{ terms})\):
\(= \lim_{x \to 1} \left[ \frac{(x + x^2 + x^3 + \dots + x^n) - (1 + 1 + 1 + \dots + 1)}{x - 1} \right]\)
Group the terms pairwise:
\(= \lim_{x \to 1} \left[ \frac{(x - 1) + (x^2 - 1) + (x^3 - 1) + \dots + (x^n - 1)}{x - 1} \right]\)
Split into individual limit terms:
\(= \lim_{x \to 1} \left[ \frac{x - 1}{x - 1} \right] + \lim_{x \to 1} \left[ \frac{x^2 - 1}{x - 1} \right] + \lim_{x \to 1} \left[ \frac{x^3 - 1}{x - 1} \right] + \dots + \lim_{x \to 1} \left[ \frac{x^n - 1}{x - 1} \right]\)
Applying the standard formula \(\lim_{x \to 1} \frac{x^k - 1^k}{x - 1} = k(1)^{k-1} = k\):
\(= 1 + 2(1)^1 + 3(1)^2 + \dots + n(1)^{n-1}\)
\(= 1 + 2 + 3 + \dots + n\)
Sum of first \(n\) natural numbers:
\(= \frac{n(n + 1)}{2} \text{ ans.}\)
In simple words: Break the large polynomial into individual standard limit fractions and sum their resulting values using the arithmetic series formula.

Exam Tip: Distributing \(n\) ones across the terms allows each power of \(x\) to pair with exactly one constant.

Q.10) If \(\lim_{x \to -a} \left[ \frac{x^9 + a^9}{x + a} \right] = 9\), find the value of \(a\).
Answer: Rewrite the expression in standard form:\(\lim_{x \to -a} \left[ \frac{x^9 - (-a)^9}{x - (-a)} \right] = 9\)
Using the standard formula \(\lim_{x \to k} \frac{x^n - k^n}{x - k} = nk^{n-1}\):
\(9(-a)^{9-1} = 9\)
\(9(-a)^8 = 9\)
Divide both sides by 9:
\((-a)^8 = 1\)
Since the power is even, \((-a)^8 = a^8\):
\(a^8 = 1\)
Taking the eighth root on both sides:
\(a = \pm 1 \text{ ans.}\)
In simple words: Match the given limit equation to the standard power formula, solve the resulting algebraic equation for \(a\), and account for even powers.

Exam Tip: Be careful with negative signs when evaluating even versus odd exponents in limit formulas.