Maharashtra Board Class 11 Maths Part 1 Chapter 7 Limits 7.2 Solutions

Std 11 Maths 1 Exercise 7.2 Solutions Commerce Maths

I. Evaluate The Following Limits:

Question 1. \( \lim_{z \to 2} \left[ \frac{z^2-5z+6}{z^2-4} \right] \)
Answer:
Solution:
\( \lim_{z \to 2} \frac{z^2-5z+6}{z^2-4} \)
\( = \lim_{z \to 2} \frac{(z-3)(z-2)}{(z+2)(z-2)} \)
\[ \text{As } z \to 2, z \neq 2 \]
\[ \therefore z-2 \neq 0 \]
\( = \lim_{z \to 2} \frac{z-3}{z+2} \)
\( = \frac{2-3}{2+2} \)
\( = \frac{-1}{4} \)
In simple words: To evaluate this limit, first factorize both the numerator and denominator. Then, cancel out the common factor \((z-2)\) since \(z\) is approaching 2 but is not equal to 2, ensuring \((z-2) \neq 0\). Finally, substitute \(z=2\) into the simplified expression to find the limit.

🎯 Exam Tip: Always look for common factors in the numerator and denominator when evaluating limits of rational functions, especially when direct substitution results in an indeterminate form like \(\frac{0}{0}\). Factorization helps in simplifying the expression before substitution.

Question 2. \( \lim_{x \to -3} \left[ \frac{x+3}{x^2+4x+3} \right] \)
Answer:
Solution:
\( \lim_{x \to -3} \frac{x+3}{x^2+4x+3} \)
\( = \lim_{x \to -3} \frac{x+3}{(x+3)(x+1)} \)
\[ \text{As } x \to -3, x \neq -3 \]
\[ \therefore x+3 \neq 0 \]
\( = \lim_{x \to -3} \frac{1}{x+1} \)
\( = \frac{1}{-3+1} \)
\( = \frac{1}{-2} \)
In simple words: Factorize the denominator to find a common factor with the numerator, which is \((x+3)\). Cancel this common factor as \(x\) approaches -3 but is not equal to -3. Then, substitute \(x=-3\) into the simplified expression to determine the limit.

🎯 Exam Tip: When evaluating limits, always check for indeterminate forms first. If \(\frac{0}{0}\) occurs, algebraic manipulation like factorization is crucial to simplify the expression before direct substitution. Pay attention to the conditions under which factors can be cancelled.

Question 3. \( \lim_{y \to 0} \left[ \frac{5y^3+8y^2}{3y^4-16y^2} \right] \)
Answer:
Solution:
\( \lim_{y \to 0} \frac{5y^3+8y^2}{3y^4-16y^2} \)
\( = \lim_{y \to 0} \frac{y^2(5y+8)}{y^2(3y^2-16)} \)
\[ \text{As } y \to 0, y \neq 0 \]
\[ \therefore y^2 \neq 0 \]
\( = \lim_{y \to 0} \frac{5y+8}{3y^2-16} \)
\( = \frac{5(0)+8}{3(0)^2-16} \)
\( = \frac{8}{-16} \)
\( = \frac{-1}{2} \)
In simple words: Begin by factoring out the highest common power of \(y\) from both the numerator and the denominator. Since \(y\) approaches 0 but is not 0, the \(y^2\) terms can be cancelled. Substitute \(y=0\) into the simplified rational expression to compute the limit.

🎯 Exam Tip: For limits involving variables approaching zero, factoring out the lowest power of the variable present in all terms of the numerator and denominator is often the key to simplification. This helps remove the indeterminate form and allows for direct substitution.

Question 4. \( \lim_{x \to -2} \left[ \frac{-2x-4}{x^3+2x^2} \right] \)
Answer:
Solution:
\( \lim_{x \to -2} \frac{-2x-4}{x^3+2x^2} \)
\( = \lim_{x \to -2} \frac{-2(x+2)}{x^2(x+2)} \)
\[ \text{As } x \to -2, x \neq -2 \]
\[ \therefore x+2 \neq 0 \]
\( = \lim_{x \to -2} \frac{-2}{x^2} \)
\( = \frac{-2}{(-2)^2} \)
\( = \frac{-2}{4} \)
\( = \frac{-1}{2} \)
In simple words: Factor out common terms from the numerator and denominator; in this case, factor -2 from the numerator and \(x^2\) from the denominator. Cancel the common \((x+2)\) term since \(x\) approaches -2 but is not equal to -2. Finally, substitute \(x=-2\) into the simplified expression to evaluate the limit.

🎯 Exam Tip: Always perform factorization to identify and cancel out terms that lead to an indeterminate form (like \(\frac{0}{0}\)). This step is crucial for simplifying rational functions before substituting the limit value, ensuring a valid result.

II. Evaluate The Following Limits:

Question 1. \( \lim_{u \to 1} \left[ \frac{u^4-1}{u^3-1} \right] \)
Answer:
Solution:
\( \lim_{u \to 1} \frac{u^4-1}{u^3-1} \)
\( = \lim_{u \to 1} \frac{(u^2-1)(u^2+1)}{(u-1)(u^2+u+1)} \)
\( = \lim_{u \to 1} \frac{(u-1)(u+1)(u^2+1)}{(u-1)(u^2+u+1)} \)
\[ \text{As } u \to 1, u \neq 1 \]
\[ \therefore u-1 \neq 0 \]
\( = \lim_{u \to 1} \frac{(u+1)(u^2+1)}{u^2+u+1} \)
\( = \frac{(1+1)(1^2+1)}{1^2+1+1} \)
\( = \frac{(2)(2)}{3} \)
\( = \frac{4}{3} \)
In simple words: Factorize the numerator using the difference of squares formula (\(a^2-b^2=(a-b)(a+b)\)) twice, and the denominator using the difference of cubes formula (\(a^3-b^3=(a-b)(a^2+ab+b^2)\)). Cancel the common factor \((u-1)\) as \(u\) approaches 1 but is not equal to 1. Substitute \(u=1\) into the simplified expression to find the limit.

🎯 Exam Tip: Remember standard algebraic identities like difference of squares and cubes. These are frequently used to factorize polynomial expressions in limits, allowing for simplification and removal of indeterminate forms before substitution.

Question 2. \( \lim_{x \to 3} \left[ \frac{1}{x-3} - \frac{9x}{x^3-27} \right] \)
Answer:
Solution:
\( \lim_{x \to 3} \left[ \frac{1}{x-3} - \frac{9x}{x^3-27} \right] \)
\( = \lim_{x \to 3} \left[ \frac{1}{x-3} - \frac{9x}{(x-3)(x^2+3x+9)} \right] \)
\( = \lim_{x \to 3} \left[ \frac{x^2+3x+9-9x}{(x-3)(x^2+3x+9)} \right] \)
\( = \lim_{x \to 3} \left[ \frac{x^2-6x+9}{(x-3)(x^2+3x+9)} \right] \)
\( = \lim_{x \to 3} \left[ \frac{(x-3)^2}{(x-3)(x^2+3x+9)} \right] \)
\[ \text{As } x \to 3, x \neq 3 \]
\[ \therefore x-3 \neq 0 \]
\( = \lim_{x \to 3} \frac{x-3}{x^2+3x+9} \)
\( = \frac{3-3}{3^2+3(3)+9} \)
\( = \frac{0}{9+9+9} \)
\( = \frac{0}{27} \)
\( = 0 \)
In simple words: First, combine the two fractions by finding a common denominator, which is \((x-3)(x^2+3x+9)\). Simplify the numerator to \((x-3)^2\). Cancel the common factor \((x-3)\) as \(x\) approaches 3 but is not equal to 3. Finally, substitute \(x=3\) into the simplified expression to get the limit.

🎯 Exam Tip: When dealing with limits of expressions involving multiple fractions, combine them into a single fraction first. This often reveals opportunities for factorization and cancellation of terms causing indeterminate forms, simplifying the limit evaluation process.

Question 3. \( \lim_{x \to 2} \left[ \frac{x^3-4x^2+4x}{x^2-1} \right] \)
Answer:
Solution:
\( \lim_{x \to 2} \frac{x^3-4x^2+4x}{x^2-1} \)
\( = \lim_{x \to 2} \frac{x(x^2-4x+4)}{x^2-1} \)
\( = \lim_{x \to 2} \frac{x(x-2)^2}{x^2-1} \)
\( = \frac{2(2-2)^2}{2^2-1} \)
\( = \frac{2(0)^2}{4-1} \)
\( = \frac{2 \times 0}{3} \)
\( = \frac{0}{3} \)
\( = 0 \)
In simple words: Factor out \(x\) from the numerator to reveal a perfect square trinomial, \((x-2)^2\). The denominator is a difference of squares. Substitute \(x=2\) directly into the expression. Since the denominator is non-zero, direct substitution is valid, leading to a numerator of zero and thus a limit of zero.

🎯 Exam Tip: Always attempt direct substitution first when evaluating limits. If the result is a definite number (not an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)), that is your limit. Factorization is useful for simplification but not always necessary if direct substitution works.

III. Evaluate The Following Limits:

Question 1. \( \lim_{x \to -2} \left[ \frac{x^7+x^5+160}{x^3+8} \right] \)
Answer:
Solution:
\( \lim_{x \to -2} \frac{x^7+x^5+160}{x^3+8} \)
\( = \lim_{x \to -2} \frac{(x^7+128) + (x^5+32)}{x^3+8} \)
\( = \lim_{x \to -2} \left[ \frac{x^7-(-2)^7}{x^3-(-2)^3} + \frac{x^5-(-2)^5}{x^3-(-2)^3} \right] \)
\( = \lim_{x \to -2} \frac{x^7-(-2)^7}{x-(-2)} \cdot \frac{x-(-2)}{x^3-(-2)^3} + \lim_{x \to -2} \frac{x^5-(-2)^5}{x-(-2)} \cdot \frac{x-(-2)}{x^3-(-2)^3} \)
\[ \text{As } x \to -2, x \neq -2 \]
\[ \therefore x+2 \neq 0 \]
\[ \text{Divide Numerator and Denominator by } x+2 \]
\( = \lim_{x \to -2} \frac{x^7-(-2)^7}{x-(-2)} \cdot \lim_{x \to -2} \frac{1}{\frac{x^3-(-2)^3}{x-(-2)}} + \lim_{x \to -2} \frac{x^5-(-2)^5}{x-(-2)} \cdot \lim_{x \to -2} \frac{1}{\frac{x^3-(-2)^3}{x-(-2)}} \)
Using the formula: \( \lim_{y \to a} \frac{y^n-a^n}{y-a} = na^{n-1} \)
\( = \left( 7(-2)^{7-1} \right) \cdot \left( \frac{1}{3(-2)^{3-1}} \right) + \left( 5(-2)^{5-1} \right) \cdot \left( \frac{1}{3(-2)^{3-1}} \right) \)
\( = \frac{7(-2)^6}{3(-2)^2} + \frac{5(-2)^4}{3(-2)^2} \)
\( = \frac{7(64)}{3(4)} + \frac{5(16)}{3(4)} \)
\( = \frac{448}{12} + \frac{80}{12} \)
\( = \frac{448+80}{12} \)
\( = \frac{528}{12} \)
\( = 44 \)
In simple words: Break down the numerator into terms that can be related to the denominator's form using the sum/difference of powers. Apply the limit formula \( \lim_{y \to a} \frac{y^n-a^n}{y-a} = na^{n-1} \) by dividing and multiplying by \((x-(-2))\) for each component. Then, substitute \(n\) and \(a\) values to calculate each part and sum them up for the final limit.

🎯 Exam Tip: For limits involving sums of powers, especially when the denominator is of the form \(x^n+a^n\) or \(x^n-a^n\), consider manipulating the expression to use the standard limit formula \(\lim_{y \to a} \frac{y^n-a^n}{y-a} = na^{n-1}\). This often simplifies complex polynomial limits efficiently.

Question 2. \( \lim_{y \to \frac{1}{2}} \left[ \frac{1-8y^3}{y-4y^3} \right] \)
Answer:
Solution:
\( \lim_{y \to \frac{1}{2}} \frac{1-8y^3}{y-4y^3} \)
\( = \lim_{y \to \frac{1}{2}} \frac{(1)^3-(2y)^3}{y(1-4y^2)} \)
\( = \lim_{y \to \frac{1}{2}} \frac{(1-2y)(1+2y+4y^2)}{y(1-2y)(1+2y)} \)
\[ \text{As } y \to \frac{1}{2}, y \neq \frac{1}{2} \]
\[ \therefore 1-2y \neq 0 \]
\( = \lim_{y \to \frac{1}{2}} \frac{1+2y+4y^2}{y(1+2y)} \)
\( = \frac{1+2(\frac{1}{2})+4(\frac{1}{2})^2}{\frac{1}{2}(1+2(\frac{1}{2}))} \)
\( = \frac{1+1+4(\frac{1}{4})}{\frac{1}{2}(1+1)} \)
\( = \frac{1+1+1}{\frac{1}{2}(2)} \)
\( = \frac{3}{1} \)
\( = 3 \)
In simple words: Factorize the numerator using the difference of cubes formula (\(a^3-b^3=(a-b)(a^2+ab+b^2)\)) and the denominator by first factoring out \(y\), then using the difference of squares (\(a^2-b^2=(a-b)(a+b)\)). Cancel the common factor \((1-2y)\) since \(y\) approaches \(\frac{1}{2}\) but is not equal to \(\frac{1}{2}\). Substitute \(y=\frac{1}{2}\) into the simplified expression to find the limit.

🎯 Exam Tip: Algebraic factorization is fundamental for simplifying complex rational expressions in limits. Recognize patterns like difference of cubes and squares to break down polynomials. Ensure that cancelled terms are non-zero at the limit point, which is usually the case when direct substitution leads to \(\frac{0}{0}\).

Question 3. \( \lim_{v \to \sqrt{2}} \left[ \frac{v^2+v\sqrt{2}-4}{v^3-3v\sqrt{2}+4} \right] \)
Answer:
Solution:
Consider, Numerator: \( v^2+v\sqrt{2}-4 \)
\[ = v^2+2\sqrt{2}v-\sqrt{2}v-4 \]
\[ = v(v+2\sqrt{2})-\sqrt{2}(v+2\sqrt{2}) \]
\[ = (v+2\sqrt{2})(v-\sqrt{2}) \]
Consider, Denominator: \( v^3-3v\sqrt{2}+4 \)
The OCR text for the denominator factorization starts with `v² - 3v√2 + 4`. This is an inconsistency between the question and the solution steps provided by the OCR image itself. Adhering to verbatim extraction, I will present the provided factorization steps as they appear, even though they factor a `v^2` term instead of the `v^3` term from the original question.
\[ v^2-3v\sqrt{2}+4 = v^2-2\sqrt{2}v-\sqrt{2}v+4 \]
\[ = v(v-2\sqrt{2})-\sqrt{2}(v-2\sqrt{2}) \]
\[ = (v-2\sqrt{2})(v-\sqrt{2}) \]
Therefore,
\( \lim_{v \to \sqrt{2}} \frac{v^2+v\sqrt{2}-4}{v^2-3v\sqrt{2}+4} \)
\( = \lim_{v \to \sqrt{2}} \frac{(v+2\sqrt{2})(v-\sqrt{2})}{(v-2\sqrt{2})(v-\sqrt{2})} \)
\[ \text{As } v \to \sqrt{2}, v \neq \sqrt{2} \]
\[ \therefore v-\sqrt{2} \neq 0 \]
\( = \lim_{v \to \sqrt{2}} \frac{v+2\sqrt{2}}{v-2\sqrt{2}} \)
\( = \frac{\sqrt{2}+2\sqrt{2}}{\sqrt{2}-2\sqrt{2}} \)
\( = \frac{3\sqrt{2}}{-\sqrt{2}} \)
\( = -3 \)
In simple words: First, factorize the numerator and the denominator. The numerator \(v^2+v\sqrt{2}-4\) factors into \((v+2\sqrt{2})(v-\sqrt{2})\). The provided solution's factorization of the denominator \(v^2-3v\sqrt{2}+4\) factors into \((v-2\sqrt{2})(v-\sqrt{2})\) (note: the original question states \(v^3\), but the solution proceeds with \(v^2\)). Cancel the common factor \((v-\sqrt{2})\) and then substitute \(v=\sqrt{2}\) into the simplified expression to get the limit.

🎯 Exam Tip: When dealing with limits, careful factorization of both numerator and denominator is key, especially if direct substitution leads to an indeterminate form. Always check for algebraic identities and common factors to simplify the expression before evaluating the limit. Be meticulous with algebraic manipulations, especially with terms involving radicals.

Question 4. \( \lim_{x \to 3} \left[ \frac{x^2+2x-15}{x^2-5x+6} \right] \)
Answer:
Solution:
\( \lim_{x \to 3} \frac{x^2+2x-15}{x^2-5x+6} \)
\( = \lim_{x \to 3} \frac{(x+5)(x-3)}{(x-2)(x-3)} \)
\[ \text{As } x \to 3, x \neq 3 \]
\[ \therefore x-3 \neq 0 \]
\( = \lim_{x \to 3} \frac{x+5}{x-2} \)
\( = \frac{3+5}{3-2} \)
\( = \frac{8}{1} \)
\( = 8 \)
In simple words: Factorize both the quadratic numerator and denominator into their linear factors. Identify and cancel the common factor \((x-3)\), as \(x\) approaches 3 but is not equal to 3. Substitute \(x=3\) into the simplified rational expression to compute the limit value.

🎯 Exam Tip: When evaluating limits of rational functions where direct substitution leads to \(\frac{0}{0}\), factorization of quadratic expressions is a standard method. Ensure you correctly identify the roots to factorize, and remember that cancelled factors must be non-zero at the limit point.