Maharashtra Board Class 11 Maths Part 1 Chapter 7 Limits 7.1 Solutions

I. Evaluate The Following Limits:

Question 1. \( \lim_{x \to 3} \left[ \frac{\sqrt{x+6}}{x} \right] \)
Answer: Solution: \( \lim_{x \to 3} \frac{\sqrt{x+6}}{x} = \frac{\lim_{x \to 3} \sqrt{x+6}}{\lim_{x \to 3} x} \) \( = \frac{\sqrt{3+6}}{3} \) \( = \frac{\sqrt{9}}{3} \) \( = \frac{3}{3} \) \( = 1 \) In simple words: This problem demonstrates how to evaluate a limit by directly substituting the value of x into the function, provided the function is continuous at that point and does not result in an indeterminate form.

🎯 Exam Tip: Direct substitution is the first method to try when evaluating limits; only if it leads to an indeterminate form (like 0/0) should other techniques be used.

Question 2. \( \lim_{x \to 2} \left[ \frac{x^3 - 2^3}{x-2} \right] \)
Answer: Solution: \( \lim_{x \to 2} \frac{x^3 - 2^3}{x-2} \)
\( = 3 \cdot (2)^{3-1} \)
\( = 3 \cdot (2)^2 \)
\( = 3 \cdot 4 \)
\( = 12 \) Alternatively, using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) Here, n = 3, a = 2. So, \( \lim_{x \to 2} \frac{x^3 - 2^3}{x-2} = 3 \cdot 2^{3-1} = 3 \cdot 2^2 = 3 \cdot 4 = 12 \) The OCR had a different solution path for the expression \( (-3) \cdot (2)^{-4} \), which is incorrect for this question. It seems to have picked up content from a different problem or made a mistake. I am providing the correct solution based on the limit formula. In simple words: This limit is solved using a standard algebraic limit formula for \( \frac{x^n - a^n}{x-a} \), which simplifies the evaluation without direct factorization.

🎯 Exam Tip: Recognize and apply standard limit formulas like \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) to quickly solve such problems.

Question 3. \( \lim_{x \to 5} \left[ \frac{x^3 - 125}{x^5 - 3125} \right] \)
Answer: Solution: \( \lim_{x \to 5} \frac{x^3 - 125}{x^5 - 3125} \) \( = \lim_{x \to 5} \frac{x^3 - 5^3}{x^5 - 5^5} \) Divide numerator and denominator by \( (x-5) \): \( = \frac{\lim_{x \to 5} \frac{x^3 - 5^3}{x-5}}{\lim_{x \to 5} \frac{x^5 - 5^5}{x-5}} \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) for both numerator and denominator: Numerator: \( n=3, a=5 \implies 3 \cdot 5^{3-1} = 3 \cdot 5^2 \) Denominator: \( n=5, a=5 \implies 5 \cdot 5^{5-1} = 5 \cdot 5^4 \) \( = \frac{3 \cdot 5^2}{5 \cdot 5^4} \) \( = \frac{3 \cdot 5^2}{5^5} \) \( = \frac{3}{5^{5-2}} \) \( = \frac{3}{5^3} \) \( = \frac{3}{125} \) In simple words: This problem is solved by recognizing powers of 5 and applying the limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) to both the numerator and the denominator after dividing by \( (x-a) \).

🎯 Exam Tip: When faced with a ratio of two indeterminate forms, apply the limit formula \( \frac{n a^{n-1}}{m a^{m-1}} \) to simplify, where 'n' and 'm' are the powers in the numerator and denominator, respectively.

Question 4. If \( \lim_{x \to 1} \left[ \frac{x^4 - 1}{x-1} \right] = \lim_{x \to a} \left[ \frac{x^3 - a^3}{x-a} \right] \), find all possible values of a.
Answer: Solution: Given: \( \lim_{x \to 1} \frac{x^4 - 1}{x-1} = \lim_{x \to a} \frac{x^3 - a^3}{x-a} \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) For the Left Hand Side (LHS): \( \lim_{x \to 1} \frac{x^4 - 1^4}{x-1} \) Here \( n=4, a=1 \) LHS \( = 4 \cdot (1)^{4-1} = 4 \cdot 1^3 = 4 \cdot 1 = 4 \) For the Right Hand Side (RHS): \( \lim_{x \to a} \frac{x^3 - a^3}{x-a} \) Here \( n=3, a=a \) RHS \( = 3 \cdot (a)^{3-1} = 3a^2 \) Equating LHS and RHS: \( 4 = 3a^2 \) \( a^2 = \frac{4}{3} \) \( a = \pm \sqrt{\frac{4}{3}} \) \( a = \pm \frac{2}{\sqrt{3}} \) In simple words: By applying the standard limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) to both sides of the given equation, we derive an algebraic expression for 'a' and solve for its possible values.

🎯 Exam Tip: Be mindful of both positive and negative roots when solving for a variable from an \( a^2 = k \) form, as both can be valid solutions.

II. Evaluate The Following Limits:

Question 1. \( \lim_{x \to 7} \left[ \frac{(\sqrt[3]{x} - \sqrt[3]{7})(\sqrt{x} + \sqrt{7})}{x-7} \right] \)
Answer: Solution: \( \lim_{x \to 7} \frac{(x^{\frac{1}{3}} - 7^{\frac{1}{3}})(x^{\frac{1}{2}} + 7^{\frac{1}{2}})}{x-7} \) Rewrite the expression by separating the terms and multiplying by \( x-7 \) in the denominator: \( = \lim_{x \to 7} \frac{x^{\frac{1}{3}} - 7^{\frac{1}{3}}}{x-7} \cdot (x^{\frac{1}{2}} + 7^{\frac{1}{2}}) \) This separation is not correct. The denominator \( (x-7) \) is for the whole expression. Let's analyze the expression. The problem seems to be intended for a different kind of simplification or formula. Given the standard pattern, the term \( (x^{\frac{1}{2}} + 7^{\frac{1}{2}}) \) in the numerator suggests it should cancel with something or be directly evaluated. Let's consider the form \( \lim_{x \to a} \frac{x^n - a^n}{x-a} \). The expression is \( \lim_{x \to 7} \frac{(x^{\frac{1}{3}} - 7^{\frac{1}{3}})(x^{\frac{1}{2}} + 7^{\frac{1}{2}})}{x-7} \) Let's separate it into two limits: \( = \lim_{x \to 7} \frac{x^{\frac{1}{3}} - 7^{\frac{1}{3}}}{x-7} \cdot \lim_{x \to 7} (x^{\frac{1}{2}} + 7^{\frac{1}{2}}) \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) for the first part: Here \( n = \frac{1}{3} \), \( a = 7 \) First part \( = \frac{1}{3} (7)^{\frac{1}{3}-1} = \frac{1}{3} (7)^{-\frac{2}{3}} \) For the second part (direct substitution): Second part \( = (7^{\frac{1}{2}} + 7^{\frac{1}{2}}) = 2 \cdot 7^{\frac{1}{2}} \) Now, multiply the two parts: \( = \left( \frac{1}{3} (7)^{-\frac{2}{3}} \right) \cdot \left( 2 \cdot 7^{\frac{1}{2}} \right) \) \( = \frac{2}{3} \cdot 7^{-\frac{2}{3} + \frac{1}{2}} \) \( = \frac{2}{3} \cdot 7^{\frac{-4+3}{6}} \) \( = \frac{2}{3} \cdot 7^{-\frac{1}{6}} \) \( = \frac{2}{3 \cdot 7^{\frac{1}{6}}} \) The provided solution in the OCR extract: \( = \lim_{x \to 7} \frac{(x^{\frac{1}{3}} - 7^{\frac{1}{3}})(x^{\frac{1}{3}} + 7^{\frac{1}{3}})}{x-7} \) -- This step is incorrect. It should be \( (\sqrt[3]{x} - \sqrt[3]{7})(\sqrt{x} + \sqrt{7}) \). The provided solution shows \( (x^{1/3} - 7^{1/3})(x^{1/3} + 7^{1/3}) \) which is `a^2 - b^2` rule and is not what is given in the question. Let's re-evaluate based on the provided OCR math: \( \lim_{x \to 7} \frac{(x^{\frac{1}{3}}-7^{\frac{1}{3}})(x^{\frac{1}{2}}+7^{\frac{1}{2}})}{x-7} \) The solution provided seems to have made an error in copying the square root/cube root part. It is assuming `(a-b)(a+b)=a^2-b^2` but the terms are `(x^1/3 - 7^1/3)` and `(x^1/2 + 7^1/2)`. Let's follow the OCR's assumed interpretation, even if it's based on an incorrect transcription, to align with its provided `Answer`. If it *were* \( (x^{1/3} - 7^{1/3})(x^{1/3} + 7^{1/3}) \), then it would be \( (x^{2/3} - 7^{2/3}) \). But the question is clearly \( (\sqrt[3]{x} - \sqrt[3]{7})(\sqrt{x} + \sqrt{7}) \). Let's stick to the actual question text and correct the solution path. The correct calculation is as derived above: \( \frac{2}{3 \cdot 7^{\frac{1}{6}}} \) However, the OCR's steps for Question 1 under II lead to: \( \frac{2}{3 \cdot 7^{\frac{1}{6}}} \) based on correct application. The OCR solution's last step: \( = \frac{2}{3 \cdot 3\sqrt{7}} \) which is \( \frac{2}{3 \cdot 7^{1/3}} \). This means it performed \( \frac{1}{3} (7)^{-2/3} \cdot (7^{1/3} + 7^{1/3}) \). No, the calculation is `(1/3) * (7^(-2/3)) * (2 * 7^(1/2))`. The OCR solution steps are: \( \lim_{x \to 7} \frac{(x^{\frac{1}{3}}-7^{\frac{1}{3}})(x^{\frac{1}{2}}+7^{\frac{1}{2}})}{x-7} \) \( = \lim_{x \to 7} \frac{(x^{\frac{1}{3}}-7^{\frac{1}{3}})}{x-7} \cdot (x^{\frac{1}{2}}+7^{\frac{1}{2}}) \) Using \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) for the first term: \( \frac{1}{3} (7)^{\frac{1}{3}-1} = \frac{1}{3} (7)^{-\frac{2}{3}} \) Substituting \( x=7 \) into the second term: \( 7^{\frac{1}{2}} + 7^{\frac{1}{2}} = 2 \cdot 7^{\frac{1}{2}} \) Product \( = \frac{1}{3} (7)^{-\frac{2}{3}} \cdot 2 \cdot 7^{\frac{1}{2}} \) \( = \frac{2}{3} \cdot 7^{-\frac{2}{3} + \frac{1}{2}} \) \( = \frac{2}{3} \cdot 7^{\frac{-4+3}{6}} \) \( = \frac{2}{3} \cdot 7^{-\frac{1}{6}} \) \( = \frac{2}{3 \cdot 7^{\frac{1}{6}}} \) The OCR then shows a result of \( \frac{2}{3 \cdot \sqrt[3]{7}} \). This is \( \frac{2}{3 \cdot 7^{1/3}} \). This is mathematically inconsistent with the preceding calculation. The final numerical output `2 / 3(3√7)` in the OCR is an error in transcription or calculation. I will provide the correct answer based on the correct mathematical derivation of the question. Corrected solution based on the question: \( \lim_{x \to 7} \frac{(\sqrt[3]{x} - \sqrt[3]{7})(\sqrt{x} + \sqrt{7})}{x-7} \) \( = \lim_{x \to 7} \frac{x^{1/3} - 7^{1/3}}{x-7} \cdot \lim_{x \to 7} (x^{1/2} + 7^{1/2}) \) First part: \( \lim_{x \to 7} \frac{x^{1/3} - 7^{1/3}}{x-7} = \frac{1}{3} (7)^{1/3 - 1} = \frac{1}{3} (7)^{-2/3} \) Second part: \( \lim_{x \to 7} (x^{1/2} + 7^{1/2}) = (7^{1/2} + 7^{1/2}) = 2 \cdot 7^{1/2} \) Multiply the two parts: \( = \frac{1}{3} (7)^{-2/3} \cdot 2 \cdot 7^{1/2} \) \( = \frac{2}{3} \cdot 7^{(-2/3) + (1/2)} \) \( = \frac{2}{3} \cdot 7^{(-4+3)/6} \) \( = \frac{2}{3} \cdot 7^{-1/6} \) \( = \frac{2}{3 \cdot 7^{1/6}} \) In simple words: This limit involves a product of terms. We separate the limit into two parts: one using the standard \( n a^{n-1} \) formula and the other using direct substitution, then multiply the results.

🎯 Exam Tip: When evaluating limits of products, try to separate them into individual limits if each part converges, or if one part simplifies using a known formula and the other allows direct substitution.

Question 2. If \( \lim_{x \to 5} \left[ \frac{x^k - 5^k}{x-5} \right] = 500 \), find all possible values of k.
Answer: Solution: Given: \( \lim_{x \to 5} \frac{x^k - 5^k}{x-5} = 500 \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) Here \( n=k, a=5 \) So, \( k \cdot 5^{k-1} = 500 \) We need to find an integer k such that this equation holds. Let's try some integer values for k. If k=1, \( 1 \cdot 5^{1-1} = 1 \cdot 5^0 = 1 \cdot 1 = 1 \neq 500 \) If k=2, \( 2 \cdot 5^{2-1} = 2 \cdot 5^1 = 10 \neq 500 \) If k=3, \( 3 \cdot 5^{3-1} = 3 \cdot 5^2 = 3 \cdot 25 = 75 \neq 500 \) If k=4, \( 4 \cdot 5^{4-1} = 4 \cdot 5^3 = 4 \cdot 125 = 500 \) So, k=4 is a solution. Comparing both sides, we get k = 4 In simple words: By applying the standard limit formula and then solving the resulting exponential equation, we found the integer value of k that satisfies the given condition.

🎯 Exam Tip: For equations involving powers and their bases, testing integer values is a common strategy when the variable is in the exponent, especially if the target number is a product of small primes.

Question 3. \( \lim_{x \to 0} \left[ \frac{(1-x)^8 - 1}{(1-x)^2 - 1} \right] \)
Answer: Solution: Let \( y = 1-x \) As \( x \to 0 \), \( y \to 1-0 \implies y \to 1 \) The limit becomes: \( \lim_{y \to 1} \frac{y^8 - 1}{y^2 - 1} \) Divide numerator and denominator by \( (y-1) \): \( = \frac{\lim_{y \to 1} \frac{y^8 - 1^8}{y-1}}{\lim_{y \to 1} \frac{y^2 - 1^2}{y-1}} \) Using the formula \( \lim_{y \to a} \frac{y^n - a^n}{y-a} = n a^{n-1} \) Numerator: \( n=8, a=1 \implies 8 \cdot 1^{8-1} = 8 \cdot 1^7 = 8 \) Denominator: \( n=2, a=1 \implies 2 \cdot 1^{2-1} = 2 \cdot 1^1 = 2 \) So, the limit is \( = \frac{8}{2} = 4 \) Alternate method (as per OCR): \( \lim_{y \to 1} \frac{y^8 - 1}{y^2 - 1} \) We can factorize the numerator and denominator: Numerator: \( y^8 - 1 = (y^4 - 1)(y^4 + 1) = (y^2 - 1)(y^2 + 1)(y^4 + 1) \) Denominator: \( y^2 - 1 \) So, \( \lim_{y \to 1} \frac{(y^2 - 1)(y^2 + 1)(y^4 + 1)}{y^2 - 1} \) Since \( y \to 1 \), \( y \neq 1 \), so \( y^2 - 1 \neq 0 \). We can cancel \( (y^2 - 1) \). \( = \lim_{y \to 1} (y^2 + 1)(y^4 + 1) \) Now, substitute \( y=1 \): \( = (1^2 + 1)(1^4 + 1) \) \( = (1 + 1)(1 + 1) \) \( = (2)(2) \) \( = 4 \) In simple words: This limit is solved by a substitution to simplify the base expression, then applying the standard limit formula for \( \frac{x^n - a^n}{x-a} \) or by algebraic factorization.

🎯 Exam Tip: For expressions involving `(1-x)^n - 1`, a substitution like `y = 1-x` often transforms the limit into a standard form that is easier to evaluate.

III. Evaluate The Following Limits:

Question 1. \( \lim_{x \to 0} \left[ \frac{\sqrt[3]{1+x} - \sqrt{1+x}}{x} \right] \)
Answer: Solution: Let \( y = 1+x \) As \( x \to 0 \), \( y \to 1+0 \implies y \to 1 \) The limit becomes: \( \lim_{y \to 1} \frac{y^{1/3} - y^{1/2}}{y-1} \) Separate into two limits: \( = \lim_{y \to 1} \left( \frac{y^{1/3} - 1}{y-1} - \frac{y^{1/2} - 1}{y-1} \right) \) -- Note: We subtracted and added 1 in the numerator implicitly: \( \frac{y^{1/3} - 1 - (y^{1/2} - 1)}{y-1} \) Applying the formula \( \lim_{y \to a} \frac{y^n - a^n}{y-a} = n a^{n-1} \) for both parts: For the first term \( \frac{y^{1/3} - 1}{y-1} \): \( n = \frac{1}{3}, a = 1 \implies \frac{1}{3} (1)^{\frac{1}{3}-1} = \frac{1}{3} \) For the second term \( \frac{y^{1/2} - 1}{y-1} \): \( n = \frac{1}{2}, a = 1 \implies \frac{1}{2} (1)^{\frac{1}{2}-1} = \frac{1}{2} \) So, the limit is \( = \frac{1}{3} - \frac{1}{2} \) \( = \frac{2 - 3}{6} \) \( = -\frac{1}{6} \) In simple words: This limit is solved by a substitution to simplify the base, then rewriting the expression as a difference of two limits, each evaluable by the standard \( n a^{n-1} \) formula.

🎯 Exam Tip: When the numerator contains multiple fractional powers, try to split the expression into separate limits by adding and subtracting '1' (or 'a' if 'a' is not 1) in the numerator, to align with the standard limit formula structure.

Question 2. \( \lim_{y \to 1} \left[ \frac{2y-2}{\sqrt[3]{7+y} - 2} \right] \)
Answer: Solution: \( \lim_{y \to 1} \frac{2y-2}{\sqrt[3]{7+y} - 2} \) Factor out 2 from the numerator: \( = \lim_{y \to 1} \frac{2(y-1)}{\sqrt[3]{7+y} - 2} \) Let \( x = 7+y \). As \( y \to 1 \), \( x \to 7+1 \implies x \to 8 \) Also, \( y-1 = x-7-1 = x-8 \) The expression becomes: \( = \lim_{x \to 8} \frac{2(x-8)}{x^{1/3} - 2} \) This is in the form \( \frac{2(x-a)}{x^n - a^n} \), which is \( 2 \cdot \frac{1}{\lim_{x \to a} \frac{x^n - a^n}{x-a}} \) Here, \( a=8 \), \( n=\frac{1}{3} \). Note that \( 2 = \sqrt[3]{8} \). \( = 2 \cdot \frac{1}{\lim_{x \to 8} \frac{x^{1/3} - 8^{1/3}}{x-8}} \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) for the denominator: \( = 2 \cdot \frac{1}{\frac{1}{3} (8)^{\frac{1}{3}-1}} \) \( = 2 \cdot \frac{1}{\frac{1}{3} (8)^{-\frac{2}{3}}} \) \( = 2 \cdot \frac{1}{\frac{1}{3} ( (2^3)^{-\frac{2}{3}} )} \) \( = 2 \cdot \frac{1}{\frac{1}{3} (2^{-2})} \) \( = 2 \cdot \frac{1}{\frac{1}{3} \cdot \frac{1}{4}} \) \( = 2 \cdot \frac{1}{\frac{1}{12}} \) \( = 2 \cdot 12 \) \( = 24 \) In simple words: This limit is evaluated by making a clever substitution to transform it into the standard \( \frac{x-a}{x^n - a^n} \) form, which is the reciprocal of the \( n a^{n-1} \) limit formula.

🎯 Exam Tip: When the denominator or numerator is of the form \( \sqrt[k]{A+y} - B \), look for a substitution \( x = A+y \) and ensure B can be expressed as \( \sqrt[k]{x} \) at the limit point.

Question 3. \( \lim_{z \to a} \left[ \frac{(z+2)^{3/2} - (a+2)^{3/2}}{z-a} \right] \)
Answer: Solution: Let \( y = z+2 \) and \( b = a+2 \) As \( z \to a \), then \( z+2 \to a+2 \implies y \to b \) The limit becomes: \( \lim_{y \to b} \frac{y^{3/2} - b^{3/2}}{ (y-2) - (b-2) } \) \( = \lim_{y \to b} \frac{y^{3/2} - b^{3/2}}{y-b} \) This is directly in the form \( \lim_{y \to b} \frac{y^n - b^n}{y-b} = n b^{n-1} \) Here \( n = \frac{3}{2} \), and the variable is \( y \) approaching \( b \). So, the limit is \( = \frac{3}{2} (b)^{\frac{3}{2}-1} \) \( = \frac{3}{2} (b)^{\frac{1}{2}} \) Substitute back \( b = a+2 \): \( = \frac{3}{2} (a+2)^{\frac{1}{2}} \) \( = \frac{3}{2} \sqrt{a+2} \) In simple words: By performing a substitution to simplify the base of the exponential terms, the limit is transformed into a direct application of the standard \( n a^{n-1} \) limit formula.

🎯 Exam Tip: Look for opportunities to simplify complex limit expressions involving sums or differences in the base by making appropriate substitutions that transform them into the standard \( \frac{x^n - a^n}{x-a} \) form.

Question 4. \( \lim_{x \to 5} \left[ \frac{x^3 - 125}{x^2 - 25} \right] \)
Answer: Solution: \( \lim_{x \to 5} \frac{x^3 - 125}{x^2 - 25} \) Recognize \( 125 = 5^3 \) and \( 25 = 5^2 \): \( = \lim_{x \to 5} \frac{x^3 - 5^3}{x^2 - 5^2} \) To apply the limit formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} \), divide both numerator and denominator by \( (x-5) \): \( = \frac{\lim_{x \to 5} \frac{x^3 - 5^3}{x-5}}{\lim_{x \to 5} \frac{x^2 - 5^2}{x-5}} \) Using the formula \( \lim_{x \to a} \frac{x^n - a^n}{x-a} = n a^{n-1} \) for both numerator and denominator: Numerator: \( n=3, a=5 \implies 3 \cdot 5^{3-1} = 3 \cdot 5^2 \) Denominator: \( n=2, a=5 \implies 2 \cdot 5^{2-1} = 2 \cdot 5^1 \) \( = \frac{3 \cdot 5^2}{2 \cdot 5^1} \) \( = \frac{3 \cdot 25}{2 \cdot 5} \) \( = \frac{75}{10} \) \( = \frac{15}{2} \) In simple words: This limit is resolved by identifying perfect cubes and squares, then dividing both the numerator and the denominator by \( (x-a) \) to apply the standard \( n a^{n-1} \) limit formula.

🎯 Exam Tip: When dealing with rational functions that yield an indeterminate form \( \frac{0}{0} \), factoring or applying the \( \frac{n a^{n-1}}{m a^{m-1}} \) shortcut for differences of powers is highly efficient.