Problem 118 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I (I) \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\ln \sin x}\) (II) \(\lim _{n \rightarrow \infty}\left\\{\frac{7}{10}+\frac{29}{10^{2}}+\frac{133}{10^{3}}+\ldots\right.\) \(\left.+\frac{5^{n}+2^{n}}{10^{n}}\right\\}\) (III) \(\lim _{x \rightarrow \infty} x\left[\tan ^{-1} \frac{x+1}{x+2}-\frac{\pi}{4}\right]\) (IV) \(\lim _{n \rightarrow \infty} \frac{n^{k} \sin ^{2}(n !)}{n+2}, 0

Step-by-Step Solution

Answer

I: -1 (Assumed mistake), II: D, III: C, IV: C

Step 1: Solve Exercise I

Let's find the limit of \( \lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\ln \sin x} \). Assume \( \sin x = t \). As \( x \rightarrow \frac{\pi}{2} \), \( t \rightarrow 1 \). Applying the L'Hôpital's Rule to solve the indeterminate form \( \frac{0}{0} \):\[ \lim_{t \rightarrow 1} \frac{t - t^t}{1-t+\ln t} = \lim_{t \rightarrow 1} \frac{1 - \ln t - t^t \cdot \ln t}{-1 + \frac{1}{t}} \]Applying L'Hôpital's Rule again:\[ \lim_{t \rightarrow 1} \frac{- \frac{1}{t} - t^t \cdot (\ln t)^2 - t^t \cdot \frac{1}{t}}{-\frac{1}{t^2}} = -1 \]

Step 2: Solve Exercise II

The sequence is a geometric series which can be formulated as:\[ \sum_{n=1}^{\infty} \frac{5^n + 2^n}{10^n} \]This can be split into two separate geometric series:\[ \sum_{n=1}^{\infty} \left( \frac{5}{10} \right)^n + \sum_{n=1}^{\infty} \left( \frac{2}{10} \right)^n \]Both are convergent since \( \left| \frac{5}{10} \right| < 1 \) and \( \left| \frac{2}{10} \right| < 1 \). The sum of each series is calculated by the formula \( \frac{a}{1-r} \), where \( a \) is the first term and \( r \) is the common ratio:\[ \frac{\frac{5}{10}}{1 - \frac{5}{10}} = 1 \quad \text{and} \quad \frac{\frac{2}{10}}{1 - \frac{2}{10}} = \frac{1}{4} \]The total sum is \( 1 + \frac{1}{4} = \frac{5}{4} \).

Step 3: Solve Exercise III

Let's find the limit of \( \lim _{x \rightarrow \infty} x\left[\tan^{-1} \frac{x+1}{x+2}-\frac{\pi}{4}\right] \).Rewrite \( \tan^{-1} \frac{x+1}{x+2} \) as \( \tan^{-1} 1 + \tan^{-1} \left[\frac{-1}{(x+2)(x+1)}\right] \) using the approximation:\[ \tan^{-1} u \approx u \quad \text{for small } u \]Therefore:\[ \lim_{x \rightarrow \infty} x \cdot \frac{-1}{(x+2)(x+1)} = \lim_{x \rightarrow \infty} \frac{-x}{x^2 + 3x + 2} \to 0 \]

Step 4: Solve Exercise IV

We need to analyze \( \lim _{n \rightarrow \infty} \frac{n^{k} \sin^{2}(n!)}{n+2} \) where \( 0 < k < 1 \).We know \( \sin^2(n!) \) is bounded between 0 and 1. Hence:\[ 0 \leq n^k \sin^2(n!) \leq n^k \]As \( n \rightarrow \infty \), \( n^k/(n+2) \rightarrow 0 \) because \( k < 1 \).Thus:\[ \lim_{n \rightarrow \infty} \frac{n^k \sin^2(n!)}{n+2} = 0 \]

Step 5: Final Matching

Match each expression from Column I to its result in Column II:- (I) matches with None (-1 is not available in Column II)- (II) matches with (D) \( \frac{5}{4} \)- (III) matches with (C) 0- (IV) matches with (C) 0

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool for evaluating indeterminate limits like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). When a limit of the form \( \frac{f(x)}{g(x)} \) results in an indeterminate form as \( x \rightarrow a \), L'Hôpital's Rule allows you to differentiate the numerator and the denominator separately and then take the limit again:

• If \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided the limit on the right exists.

In the problem **Exercise I**, we applied L'Hôpital's Rule twice to break down the complex expression \( \frac{t - t^t}{1-t+\ln t} \) into simpler components until it yielded a limit of \(-1\). By converting trigonometric functions and logarithms using differentiation, we were able to solve a seemingly complicated limit problem in a straightforward way. This method is particularly helpful when direct substitution leads to an undefined result.
Remember, the main goal is to convert the indeterminate form into something calculable.

Infinite Series

An infinite series is essentially an endless sum of terms. In mathematics, we use limits to determine whether an infinite series converges (approximates towards a finite number) or diverges (increases indefinitely). There are various tests to gauge convergence, such as the ratio test, root test, and integral test.
A geometric series is a special type of series defined by a constant ratio \( r \) between consecutive terms. The sum \( S \) of a geometric series is given by:

• \( S = \frac{a}{1-r} \) for \( |r| < 1 \)

Where \( a \) is the initial term, and \( r \) is the common ratio. If \( |r| \geq 1 \), the series does not converge.
In **Exercise II**, we identified the series as two separate geometric series with ratios \( \frac{5}{10} \) and \( \frac{2}{10} \), both of which are less than 1, thereby confirming their convergence. Calculating their individual sums and adding them gave us the total sum \( \frac{5}{4} \), a simple and elegant solution thanks to the powerful properties of geometric series.

Geometric Series

Geometric series are a type of infinite series where each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio \( r \). The classic form of a geometric series is \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term.

• If the absolute value of \( r \) is less than 1, the series converges.
• If \( |r| \geq 1 \), the series diverges.

For **Exercise II**, we split the given infinite series into two separate geometric series, each having a \( |r| < 1 \), i.e., \( \left(\frac{5}{10}\right)^n \) and \( \left(\frac{2}{10}\right)^n \). The convergence of these series allowed us to apply the formula for the sum of a geometric series. The step-by-step approach of identifying the terms, examining the ratios, and computing individual sums illustrates how geometric series can simplify complex problems elegantly. Each convergent series contributes to the overall sum of \( \frac{5}{4} \), showing how pattern recognition and the application of known formulas facilitate finding solutions efficiently.