Problem 122 - A Complete Resource Book in Mathematics for JEE Main

Question

Instructions: In the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: \(\lim _{x \rightarrow 0} \frac{e-(1+x)^{1 / x}}{x}=\frac{e}{2}\) Reason: \(\lim _{x \rightarrow 0} \frac{\ln (1+x)-x}{x}=0\) and \(\lim _{x \rightarrow 0} \frac{\ln (1+x)-x}{x^{2}}=-\frac{1}{2}\)

Step-by-Step Solution

Answer

B: Both are true, but the reason does not explain the assertion.

Step 1: Evaluate the Assertion

The Assertion is given as \( \lim_{x \to 0} \frac{e - (1 + x)^{1/x}}{x} = \frac{e}{2} \).To solve this, we'll use the expansion of \((1 + x)^{1/x}\) around \(x = 0\), i.e., \((1 + x)^{1/x} \approx e^{ (1/x)\ln(1+x) }\). For small \(x\), \(\ln(1+x) \approx x - \frac{x^{2}}{2} + \ldots\) so,\[ (1 + x)^{1/x} \approx e^{x - \frac{x^{2}}{2}} \approx e \left(1 - \frac{x}{2}\right) \]Therefore, \[ e - (1 + x)^{1/x} \approx e - e \left(1 - \frac{x}{2}\right) = e \cdot \frac{x}{2} \]Thus,\[ \lim_{x \to 0} \frac{e - (1 + x)^{1/x}}{x} = \lim_{x \to 0} \frac{e \cdot \frac{x}{2}}{x} = \frac{e}{2} \]Therefore, the Assertion is true.

Step 2: Evaluate the Reason

The Reason is given as two separate limits:1. \( \lim_{x \to 0} \frac{\ln (1 + x) - x}{x} = 0 \)2. \( \lim_{x \to 0} \frac{\ln (1 + x) - x}{x^{2}} = -\frac{1}{2} \)For the first limit:Using the expansion \(\ln(1 + x) \approx x - \frac{x^2}{2} \), we have:\[ \ln(1+x) - x \approx -\frac{x^2}{2} \]So,\[ \lim_{x \to 0} \frac{\ln (1 + x) - x}{x} = \lim_{x \to 0} \frac{-\frac{x^2}{2}}{x} = \lim_{x \to 0} -\frac{x}{2} = 0 \]This statement is true.For the second limit:\[ \lim_{x \to 0} \frac{\ln(1+x) - x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2}}{x^2} = -\frac{1}{2} \]This statement is also true.Therefore, the Reason is true.

Step 3: Determine the Relationship

Both the Assertion and the Reason are true. However, the Reason does not explain the Assertion directly. The assertion is about evaluating a limit that involves \((1+x)^{1/x}\), not directly related to the specific form of the limits given in the Reason.Thus, Reason (R) is not a correct explanation for Assertion (A).

Key Concepts

L'Hôpital's Rule

When dealing with limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule is a powerful tool to use. This rule allows you to differentiate the numerator and the denominator separately and then take the limit again. This is especially handy when solving limits that seem complex or daunting.

To use L'Hôpital’s Rule effectively, make sure that

• the original limit fits the criteria of being an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• the functions in the numerator and the denominator are differentiable.
• you can apply L'Hôpital's Rule again if the result is still indeterminate.

Once these criteria are met,
you differentiate the numerator and the denominator,
then test the new limit. Using this method simplifies the process of finding more complex limits.

Maclaurin Series

The Maclaurin Series is a specific type of Taylor series that approximates functions at zero. It's particularly useful when we're dealing with exponential functions, sine, cosine, and logarithmic functions, as it's a way to express them as an infinite sum of terms.

For example, the function \( e^x \) can be expressed as:
\[ e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

Using a Maclaurin Series allows us to break down complex functions to easier-to-manage polynomial expressions. In the context of our problem,
we used the Maclaurin expansion of \( \ln(1 + x) \), which is \( x - \frac{x^2}{2} + \ldots \).
This approximation helped in simplifying and making calculations easier when evaluating limits.

Indeterminate Forms

Indeterminate forms arise in calculus when the limit of a function is not immediately clear. Common examples include \( \frac{0}{0} \), \( \infty - \infty \), and \( 0^0 \). These forms signal that more advanced techniques are required to find the limit value.

In the exercise given,
both the assertion and reason deal with indeterminate forms. The trick to managing these forms lies in applying calculus methods like L'Hôpital's Rule, algebraic manipulation, or series expansions to clarify and resolve the form into something that is determinable.

• First, identify that you indeed have an indeterminate form;
• then choose an appropriate method to resolve it.

For example, using a Taylor or Maclaurin series can transform a troublesome expression into a polynomial which is much easier to evaluate. Indeterminate forms may initially seem evasive, but with the right techniques, they become manageable.