Problem 118 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

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

Step-by-Step Solution

Answer

Matches: (I)-(C), (II)-(D), (III)-(C), (IV)-(B)

Step 1: Solve Expression I

For the limit \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x-(\sin x)^{\sin x}}{1-\sin x+\ln \sin x}\), first observe that as \(x \rightarrow \frac{\pi}{2}\), \(\sin x \rightarrow 1\). Thus the expression becomes an indeterminate form \(\frac{0}{0}\) and can be solved using L'Hôpital's rule. Differentiate the numerator and the denominator. The derivative of \(\sin x\) is \(\cos x\) and using the chain rule, the derivative of \((\sin x)^{\sin x}\) is \(\sin x \cos x (\ln \sin x + 1)\). For the denominator, the derivative of \(1 - \sin x + \ln \sin x\) is \(-\cos x + \frac{1}{\sin x}\cdot\cos x\), hence simplifying \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin x - (\sin x)^{\sin x}}{1-\sin x+\ln \sin x}\) yields \(0\). So this matches (C).

Step 2: Solve Expression II

Consider the limit \(\lim _{n \rightarrow \infty} \left\{ \frac{7}{10} + \frac{29}{10^2} + \frac{133}{10^3} + \cdots + \frac{5^n + 2^n}{10^n}\right\}\). This series can be rewritten using the fact that for large \(n\), the sequence simplifies to a geometric sequence dominated by \(\frac{5^n}{10^n} = \left(\frac{5}{10}\right)^n = \left(\frac{1}{2}\right)^n\). Thus the sum approaches \(\frac{5}{4}\). This matches (D).

Step 3: Solve Expression III

For the limit \(\lim _{x \rightarrow \infty} x\left[\tan^{-1} \frac{x+1}{x+2} - \frac{\pi}{4}\right]\), use approximation expressions. As \(x \rightarrow \infty\), \(\frac{x+1}{x+2} \approx 1\), so \(\tan^{-1} \frac{x+1}{x+2} \rightarrow \frac{\pi}{4}\). The first-order Taylor expansion for \(\tan^{-1} (1 + h) \approx \frac{\pi}{4} + \frac{h}{1+h^2}\) can be used. Our expression reduces to \[-x \cdot \frac{1/2}{2} = -\frac{x}{4}\] which tends to zero as \(x \rightarrow \infty\). Therefore, Expression III matches (C).

Step 4: Solve Expression IV

The expression \(\lim _{n \rightarrow \infty} \frac{n^{k} \sin^{2}(n!)}{n+2}\) provides a sine function nested in a factorial. Since \(\sin^{2}(x)\) is always between 0 and 1, as \(n\) increases, the denominator \(n+2\) grows larger faster than \(n^k\). As \(0 < k < 1\), \(n^k\) grows slower than \(n\). Therefore, the numerator becomes negligible compared to the denominator, causing the overall fraction to approach zero for large \(n\). This matches (B).

Key Concepts

L'Hôpital's Rule

When evaluating limits, you may encounter expressions that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are cases where direct substitution does not provide a clear answer.
In these situations, L'Hôpital's Rule becomes a valuable tool. It allows you to find the limit by differentiating the numerator and the denominator separately. You continue to use this rule until the expression is no longer an indeterminate form.
Recall from the previous exercise:

• Identify the indeterminate form \( \frac{0}{0} \) as \(x \to \frac{\pi}{2}\) for the given limit.
• Differentiate the numerator \(\sin x - (\sin x)^{\sin x}\).
• Differentiate the denominator \(1 - \sin x + \ln \sin x\).
• Re-evaluate the limit with these derivatives to simplify the expression, leading to a result of 0.

L'Hôpital's Rule allows you to straightforwardly tackle problems that initially seem puzzling due to indeterminate forms.

Indeterminate Forms

Indeterminate forms arise when the direct application of a limit results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others such as \( 0 \times \infty \), \( \infty - \infty \). These are scenarios that signify uncertainty or ambiguity in calculations.
More advanced techniques, like L'Hôpital's Rule or algebraic manipulation, can handle these situations effectively.
Here's what to consider:

• Indeterminate forms imply there is a special condition in the problem that requires extra analysis.
• The expression \( \sin x - (\sin x)^{\sin x} \) near \( x = \frac{\pi}{2} \) turns into \( 0 \) as both components reach 1.
• This situation called for recognizing a \( \frac{0}{0} \) case and allowed the application of L'Hôpital’s Rule in the exercise.
• Focus on transforming expressions first to check if there are simplifications possible to resolve the indeterminacy without derivatives.

Understanding and recognizing indeterminate forms can help you choose appropriate strategies to evaluate complex calculus problems.

Infinite Series

Infinite series extend the idea of summation to an infinite number of terms. Learning how they work is one of the foundations of calculus because they help describe functions in terms of sums.
In the previous exercise, an infinite series like this was used:

• Rewriting the limit to recognize it as a geometric series problem.
• Understanding that for large \(n\), the sequence simplifies so that \( \left(\frac{5}{10}\right)^n \) dominates.
• Evaluating the series using the formula for infinite geometric series even though it initially looks complex.

Infinite series help you grasp more complex functions through representation as infinite sums and are crucial for understanding a range of mathematical applications.