Problem 49 - A Complete Resource Book in Mathematics for JEE Main

Question

$$ \lim _{x \rightarrow 0} \frac{x \sqrt[3]{z^{2}-(z-x)^{2}}}{\left(\sqrt[3]{8 x z-4 x^{2}}+\sqrt[3]{8 x z}\right)^{4}} \text { is equal to } $$ (A) \(\frac{z}{2^{11 / 3}}\) (B) \(\frac{1}{2^{23 / 3} \cdot z}\) (C) \(2^{21 / 3} z\) (D) None of these

Step-by-Step Solution

Answer

The answer is (D) None of these.

Step 1: Simplify the Expression

First, rewrite the denominator using the common cube root and simplify each term using an approximation for small \( x \):\[ \text{Denominator: } \left( \sqrt[3]{8xz - 4x^2} + \sqrt[3]{8xz} \right)^4 \]For small \( x \), \( \sqrt[3]{8xz - 4x^2} \approx \sqrt[3]{8xz} \), hence the expression simplifies to twice \( \sqrt[3]{8xz} \) in the cube root. This simplifies further:\[ \left( 2 \cdot \sqrt[3]{8xz} \right)^4 = 16 \cdot \left( 8xz \right)^{4/3} = 16 \cdot 8^{4/3} \cdot x^{4/3} \cdot z^{4/3} \]

Step 2: Simplify the Numerator

Now, simplify the numerator:\[ x \cdot \sqrt[3]{z^2 - (z-x)^2} \]Expand the expression inside the cube root:\[ z^2 - (z-x)^2 = z^2 - (z^2 - 2zx + x^2) = 2zx - x^2 \]For small \( x \), \( \sqrt[3]{2zx - x^2} \approx \sqrt[3]{2zx} \).Thus, \( x \cdot \sqrt[3]{2zx - x^2} \approx x \cdot \sqrt[3]{2zx} \).

Step 3: Form the Fraction and Simplify

Put the simplified expressions into the limit expression:\[ \frac{x \cdot \sqrt[3]{2zx}}{16 \cdot 8^{4/3} \cdot x^{4/3} \cdot z^{4/3}} \]Cancel the common \( x \) terms:\[ \frac{x^{1 - 4/3}}{16 \cdot 8^{4/3}} \cdot \frac{(2zx)^{1/3}}{z^{4/3}} \] which simplifies to\[ \frac{x^{-1/3} \cdot 2^{1/3} \cdot z^{1/3}}{16 \cdot 8^{4/3}} \].

Step 4: Evaluate the Limit as x Approaches Zero

Since \( x^{-1/3} \) approaches infinity as \( x \) approaches zero, rewrite the expression in terms of constants and evaluate:\[ \frac{2^{1/3} \cdot z^{1/3}}{16 \cdot 8^{4/3}} = \frac{z^{1/3}}{16 \cdot 2^{11/3}} \]The limit evaluates to zero, indicating the expression is dominated by the denominator.

Step 5: Confirm the Simplification Fits Options

Recognize that as \( x \to 0 \), `Step 4` simplified expression approaches zero because of the \( x^{-1/3} \) term and the common constant values. Thus it matches none of the options that resolve into real numbers with non-zero evaluations without \( x \).

Key Concepts

Limit Evaluation Techniques

In calculus, evaluating limits is a crucial skill, especially when dealing with complex expressions. A variety of limit evaluation techniques exist to facilitate solving these problems effectively. Here are some common strategies often used to simplify the limit problems:

• **Direct Substitution**: The simplest method, applicable when you can directly substitute the value of the variable approaching a certain number.
• **Factoring**: If direct substitution leads to an indeterminate form, try factoring the expression to simplify it.
• **Rationalizing**: This technique involves multiplying the numerator and denominator by conjugates to simplify the expression.
• **Common Denominators**: Simplifying by combining fractions using a common denominator can also be helpful.
• **Series Expansions**: Applying Taylor or Maclaurin series expansions can transform complex functions into manageable polynomial forms.
• **Variable Change**: Sometimes substituting another variable can simplify the expression and reveal the limit.

In the given exercise, these techniques were combined. By using approximations for small values of variables and algebraic manipulation, the expressions were simplified before evaluating the limit.

Indeterminate Forms

Indeterminate forms appear in limit problems when substitution of the approaching value does not lead to an immediate answer or results in undefined forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Recognizing these forms is crucial because it allows you to apply special rules or techniques to find the limit successfully.
When facing an indeterminate form:

• Look for opportunities to factor or cancel out terms.
• Consider transformations that might simplify the expression, such as splitting or combining terms.
• Apply algebraic modifications like conjugates or polynomial expansions.

In the exercise, the division initially led to such an indeterminate form as certain terms in the numerator and denominator both approached zero. Simplifying by factoring and approximating near small values allowed for manageable calculations that resolved the indeterminate scenario into a more solvable form.

L'Hôpital's Rule

L'Hôpital's Rule is an elegant tool for resolving limits that initially result in indeterminate forms. It provides a formula to differentiate the numerator and the denominator independently and then compute the limit, if other methods fail or are too complex. The rule applies under these conditions:

• The limit leads to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Both numerator and denominator are differentiable around the point of interest.
• The new limit does not again resolve into an indeterminate form infinitely.

While the original exercise did not ultimately require L'Hôpital's Rule, as approximations and other simplifications sufficed, it's a reliable fallback. It epitomizes a methodical approach to resolving tricky limits where repeated differentiation can bring clarity and resolution, especially in more complex scenarios where algebraic simplifications alone might not work.