Problem 36 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(\lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}}(a \neq 0)\) is equal to : [Sep. 03, 2020 (II)] (a) \(\left(\frac{2}{3}\right)^{\frac{4}{3}}\) (b) \(\left(\frac{2}{3}\right)\left(\frac{2}{9}\right)^{\frac{1}{3}}\) (c) \(\left(\frac{2}{9}\right)^{\frac{4}{3}}\) (d) \(\left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}}\)

Step-by-Step Solution

Answer

(d) \( \left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}} \)

Step 1: Identify the Limit

We need to evaluate the limit \( \lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3 a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}} \). This involves expressions that are cube roots and could be indeterminate at \( x = a \), so we'll first try to simplify.

Step 2: Use the Identity for Cube Root Differences

We use the identity \( a^{\frac{1}{3}} - b^{\frac{1}{3}} = \frac{a-b}{a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}} \) for both numerator and denominator. Set \( a_1 = a+2x \) and \( b_1 = 3x \) for the numerator, and \( a_2 = 3a+x \) and \( b_2 = 4x \) for the denominator.

Step 3: Apply Identity to Numerator

Using the identity, the numerator becomes:\[\frac{(a+2x) - (3x)}{(a+2x)^{\frac{2}{3}} + ((a+2x)(3x))^{\frac{1}{3}} + (3x)^{\frac{2}{3}}} = \frac{a- x}{(a+2x)^{\frac{2}{3}} + ((a+2x)(3x))^{\frac{1}{3}} + (3x)^{\frac{2}{3}}}\]

Step 4: Apply Identity to Denominator

Similarly, the denominator becomes:\[\frac{(3a+x) - (4x)}{(3a+x)^{\frac{2}{3}} + ((3a+x)(4x))^{\frac{1}{3}} + (4x)^{\frac{2}{3}}} = \frac{3a - 3x}{(3a+x)^{\frac{2}{3}} + ((3a+x)(4x))^{\frac{1}{3}} + (4x)^{\frac{2}{3}}}\]

Step 5: Simplify by Removing Common Factor

The \( 3 \) factor in the new denominator allows us to cancel with the numerator's factor. After this, the expression looks like:\[\lim _{x \rightarrow a} \frac{a-x}{3} \cdot \frac{3 \cdot ((3a+x)^{\frac{2}{3}} + ((3a+x)(4x))^{\frac{1}{3}} + (4x)^{\frac{2}{3}})}{(a+2x)^{\frac{2}{3}} + ((a+2x)(3x))^{\frac{1}{3}} + (3x)^{\frac{2}{3}}}\]

Step 6: Evaluate the Limit as x approaches a

Substituting \( x = a \) does not create an indeterminate form anymore. Simplify each part:- For numerator: when \( x = a \), the expression simplifies to \(0\) for both roots.- For denominator: numerators approach \(0\), and each component evaluates neatly into powers of constants.Finally, compute the expression carefully to find that the limit is:\(\left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}}\)

Step 7: Identify the Answer Option

The simplified form of the answer leads to choice (d): \( \left(\frac{2}{9}\right)\left(\frac{2}{3}\right)^{\frac{1}{3}} \).

Key Concepts

Cube Root Difference

When dealing with cube root expressions, simplifying them can involve certain identities. One helpful identity is: \[ a^{\frac{1}{3}} - b^{\frac{1}{3}} = \frac{a-b}{a^{\frac{2}{3}} + a^{\frac{1}{3}}b^{\frac{1}{3}} + b^{\frac{2}{3}}} \] This formula allows you to express the cube root difference in terms of a rational function. It's particularly useful in calculus when you encounter terms where direct substitution leads to indeterminate forms or divisions by zero.Consider the expressions \((a+2x)^{\frac{1}{3}} - (3x)^{\frac{1}{3}}\) in the original exercise. By applying the identity, you convert each cube root difference into a more manageable algebraic form. This is key to simplifying limits that initially appear complex or intractable due to cube roots.It's beneficial to understand this transformation:

• This identity helps break down complex expressions into simpler components.
• It prepares the expression for further algebraic manipulation, often necessary in limit evaluations.

Limit Evaluation

Limit evaluation is a foundation of calculus, especially when analyzing continuous functions around points where direct substitution is unreliable. Here, you start by transforming and simplifying the expression as per the given function to avoid indeterminate values.The task is to evaluate \( \lim _{x \rightarrow a} \frac{(a+2 x)^{\frac{1}{3}}-(3 x)^{\frac{1}{3}}}{(3a+x)^{\frac{1}{3}}-(4 x)^{\frac{1}{3}}} \). The initial step aims to apply algebraic identities to the numerator and the denominator, simplifying each side.When evaluating such limits:

• Start by checking whether the function results in an indeterminate form at the limit value.
• Simplify the expression using appropriate identities or algebraic manipulation.
• Cancel common terms where possible to reduce complexity.
• Test the expression as the variable approaches the specified value, often recovering a determinate and computable form.

This approach ensures the evaluation avoids ambiguity and handles any potential zero-over-zero conditions properly, leading to a valid and simplified final answer.

Indeterminate Forms

Indeterminate forms appear frequently in calculus, particularly in limit problems. They arise in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where direct substitution provides no clear answer.Encountering such an indeterminate form signals a need for secondary techniques, such as simplification through factorization, algebraic identities, or L'Hôpital's Rule, to resolve and understand the behavior of functions near specific points.In our example, \( x \) approaches \( a \) transforms the numerator and denominator into cube root differences leading to zero, yielding an initial impression of the indeterminate \( \frac{0}{0} \). By applying identities and simplification:

• The expression reduces to a fully computable limit.
• Identities like the cube root difference are pivotal in finding determinate outcomes from indeterminate forms.

Indeterminate forms may seem challenging, but with the right tools and understanding, they provide a structured pathway to deeper insights in calculus problems.