Problem 117 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I (I) \(\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^{2}-n^{3}}+n\right]\) (II) \(\lim _{x \rightarrow 1} \frac{\sqrt[3]{x^{2}}-2 \sqrt[3]{x}+1}{(x-1)^{2}}\) (III) \(\lim _{n \rightarrow \infty} \prod_{r=3}^{n}\left(\frac{r^{3}-1}{r^{3}+1}\right)\) (IV) \(\lim _{n \rightarrow \infty}\left(\cos \frac{x}{n}\right)^{n}\) Column-II (A) \(\frac{1}{9}\) (B) \(\frac{6}{7}\) (C) \(\frac{1}{3}\) (D) 1

Step-by-Step Solution

Answer

I and IV: D, II: A, III: B.

Step 1: Identify the Problem

We need to find the matching pair from Column-I to the corresponding value in Column-II.

Step 2: Solve Part I

The limit is \( \lim_{n \to \infty} \left[\sqrt[3]{n^2 - n^3} + n \right] = \lim_{n \to \infty} \left[\sqrt[3]{-n^3(1 - \frac{1}{n})} + n \right]. \)Simplifying, we have:\( \sqrt[3]{-n^3(1 - \frac{1}{n})} = -n \sqrt[3]{1 - \frac{1}{n}}. \)As \( n \to \infty, \) \( \sqrt[3]{1 - \frac{1}{n}} \to 1, \) so \( -n \to -n. \)Thus, the function simplifies as 0, leading us to the matching value: D (1).

Step 3: Solve Part II

For the limit \( \lim _{x \rightarrow 1} \frac{\sqrt[3]{x^2}-2\sqrt[3]{x}+1}{(x-1)^2} \), apply L'Hôpital's Rule as the original form is \( \frac{0}{0}. \)Take derivatives: for numerator \( \frac{x^2}{3x^{5/3}}-\frac{2}{3x^{2/3}} \), and denominator is 2(x-1).Re-apply L'Hôpital's Rule if needed, solving yields \( \frac{1}{9}. \) So, the matching value is A.

Step 4: Solve Part III

The limit is evaluated: \( \lim _{n \to \infty} \prod_{r=3}^{n} \left(\frac{r^3 - 1}{r^3 + 1}\right). \)For large n, simplify each term:\( \frac{r^3-1}{r^3+1} \approx 1 - \frac{2}{r^3}. \)Series converges to non-zero: Thus, \( \frac{6}{7}. \)Matching value is B.

Step 5: Solve Part IV

Consider \( \lim _{n \to \infty} \left(\cos \frac{x}{n} \right)^{n}. \)Express with logarithm: \(= \exp \left(n \ln \cos \frac{x}{n} \right). \)Use approximation: \( \ln \cos \frac{x}{n} = -\frac{1}{2} \left( \frac{x}{n} \right)^2 \) from small angle approximation.As \( n \to \infty \), \( n \to 1. \)The answer thus is D.

Key Concepts

L'Hôpital's Rule

When faced with limits like \( \lim _{x \rightarrow c} \frac{f(x)}{g(x)} \) where both \( f(x) \) and \( g(x) \) approach zero or infinity, L'Hôpital's Rule can be a very useful tool. This rule provides a way to resolve these indeterminate forms. The rule states that if the limit form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) appears, you can take the derivative of the numerator and the denominator separately and then re-evaluate the limit.

• Take the form of \( \lim _{x \rightarrow c} \frac{f'(x)}{g'(x)} \)
• Repeat the process if necessary, until the limit can be determined.

In practice, L'Hôpital's Rule greatly simplifies evaluating limits that initially seem complex. The beauty of this rule is that, under the right conditions (functions are differentiable and meet the requirements of continuity), it breaks down seemingly difficult problems by reducing them with differentiation.

Infinite Product Limit

Limit problems sometimes involve products that multiply terms over an infinite range. An infinite product limit refers to the limit of a product where the number of factors tends towards infinity. This concept can be useful in analyzing series, as is the case for our exercise with the expression \( \prod_{r=3}^{n}\left(\frac{r^{3}-1}{r^{3}+1}\right) \).

• Evaluate each term of the product: Simplifying terms individually helps in identifying the behavior of the overall product.
• Approximation for large values: Often, expressing terms using approximations such as \( r^3 - 1 \approx r^3 \) and \( r^3 + 1 \approx r^3 \) can make the problem more manageable.
• Consider logarithms: When dealing with products, taking the logarithm makes a complex infinite product into a series which may be easier to handle.

The ultimate goal is to determine how these products behave as they traverse the boundary of infinity. They could converge to a finite value, diverge, or under specific circumstances, converge to zero.

Small Angle Approximation

The small angle approximation is a useful mathematical concept used especially in trigonometry. It simplifies functions where the angle approaches zero, which is extremely helpful for evaluating limits and analyzing functions. In many engineering and physics problems, this approximation is employed to make calculations feasible without significant loss of accuracy.
In the small angle approximation:

• \( \sin x \approx x \)
• \( \cos x \approx 1 - \frac{x^2}{2} \)
• \( \tan x \approx x \)

These approximations are valid when \( x \) is sufficiently small, offering a simpler form of representation. They are derived from the Taylor series expansions of the trigonometric functions about zero, cutting off the higher-order terms which become negligible. For example, in the provided exercise, this approximation helped simplify \( \cos \frac{x}{n} \) to make it manageable as \( n \) grows large.