Problem 117 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

II matches (A) \( \frac{1}{9} \); III and IV match (D) 1.

Step 1: Simplifying Expression for Limit I

Consider \ \( \lim_{n \rightarrow \infty}\left(\sqrt[3]{n^2-n^3}+n\right) \). Rewriting \( \sqrt[3]{n^2-n^3} \) gives \( \sqrt[3]{n^3(1-n^{-1})} = n (1-n^{-1})^{1/3} \approx n(1 - \frac{1}{3n}) \text{ for large } n \). Substituting back, the expression becomes \ \( n(1 - \frac{1}{3n}) + n = n - \frac{1}{3} + n = 2n - \frac{1}{3} \). Thus, \( \lim_{n \rightarrow \infty} (2n - \frac{1}{3}) = \infty \), which does not match any option.

Step 2: Simplifying Expression for Limit II

Consider \( \lim_{x \rightarrow 1} \frac{\sqrt[3]{x^2} - 2\sqrt[3]{x} + 1}{(x-1)^2} \). Perform a Taylor expansion around \( x = 1 \). Set \( y = \sqrt[3]{x} \), then \( \text{as } x \rightarrow 1, y \rightarrow 1 \), and \( y^3 = x \). The expression becomes \( \frac{y^2 - 2y + 1}{(y^3-1)^2} \approx \frac{(y-1)^2}{(3(y-1))^2} \). Simplifying gives \( \frac{1}{9} \), matching option \( A \).

Step 3: Simplifying Expression for Limit III

Consider \( \lim_{n \rightarrow \infty} \prod_{r=3}^{n}\left(\frac{r^3-1}{r^3+1}\right) \). Notice the telescoping nature of the product where terms cancel out. Writing initial terms: \ \( \frac{3^3-1}{3^3+1} \frac{4^3-1}{4^3+1} \ldots \frac{n^3-1}{n^3+1} \). Simplifying gives a product of fractions each tending toward \( \frac{1}{1} \) as \( r \rightarrow \infty \) thus leaving \( 1 \). Therefore, the limit is \( 1 \), matching option \( D \).

Step 4: Simplifying Expression for Limit IV

Consider \( \lim_{n \rightarrow \infty} \left(\cos \frac{x}{n}\right)^n \). Use the continuity of cosine and that for small \(x/n\), \( \cos \frac{x}{n} \approx 1 - \frac{1}{2} (\frac{x}{n})^2 \). Hence, \( \left(\cos \frac{x}{n}\right)^n \approx \left(1 - \frac{x^2}{2n^2}\right)^n \). Using \( \lim_{n \rightarrow \infty} \left( 1 - \frac{a}{n} \right)^n = e^{-a} \), this becomes: \ \( \lim_{n \rightarrow \infty} e^{-\frac{x^2}{2n}} = e^0 = 1 \). Therefore, the limit is \( 1 \), matching option \( D \).

Key Concepts

Infinity Limits

Infinity limits are an essential concept in calculus and analysis, used to evaluate the behavior of a function as the input grows towards infinity. In mathematical terms, if you have a function \( f(x) \), an infinity limit examines what \( ext{lim}_{x \to \infty} f(x) \) becomes. This can either be a finite number, infinity, or not exist at all. For example, consider the expression \( \lim_{n \to \infty} (2n - \frac{1}{3}) \). Here, as \( n \to \infty \), the term \( 2n \) will dominate, leading the expression itself towards infinity, and so the result is \( \infty \). This understanding helps in identifying the benchmark or asymptotic behavior of sequences and functions, particularly useful in comparing rates of growth.
When solving problems involving infinity limits, it is useful to simplify expressions to identify dominant terms easily. Cancellation of lower order terms when compared to the highest power term is commonplace. Understanding these concepts equips you to better manage large expressions that "go rogue" at infinity, honing your skill in functional analysis.

Taylor Expansion

The Taylor Expansion is a powerful tool in mathematics that allows the approximation of functions by polynomials. This is particularly useful when exactly solving mathematical expressions is difficult or impossible. It's particularly handy near a point of interest. For example, around \( x = 1 \), you can approximate differentiable functions using their derivatives at that point.
In an exercise, consider the limit \( \lim_{x \to 1} \frac{\sqrt[3]{x^2} - 2\sqrt[3]{x} + 1}{(x-1)^2} \). Here, a Taylor expansion helps approximate the cube roots, rearranging them as simpler terms. By setting \( y = \sqrt[3]{x} \) and wondering what happens as \( y \to 1 \), it rearranges the problem into a rational function that is easier to solve, yielding \( \frac{1}{9} \).
The general formula for a Taylor series around a point \( a \) for a function \( f(x) \) is given by:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)

This approach simplifies the evaluation of complex expressions to achievable polynomial standards, thus bridging the gap between exact and practical problem-solving.

Telescoping Series

Telescoping series are special types of series where many terms cancel out the preceding or subsequent terms, leaving only a few terms remaining. These are especially useful for finding limits of sequences and series.
Consider the expression \( \lim_{n \to \infty} \prod_{r=3}^{n}\left(\frac{r^3-1}{r^3+1}\right) \). By expanding this product, observe that successive terms cancel each other out, resembling a telescope's motion closing in to reveal only essential parts. Typically, these series compress to unity or some other finite value, leading here to a limit of \( 1 \).
In these types of problems, identifying the sequence of cancellations allows the simplification of fractions into neater outcomes. Telescoping series can initially seem complex, but by recognizing the pattern of cancellations, students can solve what might look insurmountable at first.

Continuity of Functions

Continuity refers to a function being continuous at a point or over an interval, meaning it has no holes, jumps, or breaks. The function's behavior at neighboring points aligns with its activity at the point in question.
An example of practical continuity application is in the limit \( \lim_{n \to \infty} \left(\cos \frac{x}{n}\right)^n \). Here, as \( n \to \infty \), the argument of cosine becomes minuscule, nudging \( \cos(\cdot) \) toward 1. Leveraging the limits of exponential functions, this further simplifies to 1.
Continuous functions are essential in calculus since they indicate predictable behavior across their domains. This predictability allows mathematicians and learners alike to use simpler approximations consistently without sudden functional mismatches. Grasping continuity ensures a deeper understanding of the behavior of functions under loads of calculus operations.