Problem 76 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{1 / x}=\) (A) \(e^{1 / 2}\) (B) \(e^{-1 / 2}\) (C) \(e^{1 / 4}\) (D) None of these

Step-by-Step Solution

Answer

(B) \(e^{-1/2}\).

Step 1: Rewrite the Expression

Rewrite the expression in terms of a single limit. Given the function, we need to analyze it in the standard exponential form. To do this, recognize that the expression is of the form \(\lim_{x \to 0} \left(1 + \frac{f(x)}{x}\right)^{1/x}\) where \(f(x) = x - 1 + \cos x\).

Step 2: Analyze the Term Inside the Limit

We first focus on \(f(x) = x - 1 + \cos x\) and need to approximate it as \(x\) tends to zero. Expand \(\cos x\) using the Taylor series expansion around zero: \(\cos x \approx 1 - \frac{x^2}{2}\). Substitute this in to get: \(f(x) = x - 1 + (1 - \frac{x^2}{2}) = x - \frac{x^2}{2}\).

Step 3: Simplify the Expression Inside the Exponential

Substitute the expanded \(f(x)\) back into the limit expression: \(\frac{f(x)}{x} = \frac{x - \frac{x^2}{2}}{x} = 1 - \frac{x}{2}\). Therefore, the whole expression becomes: \(\lim_{x \to 0} \left(1 + \frac{1 - \frac{x}{2}}{x}\right)^{1/x}\).

Step 4: Simplify the Exponential Form

Recognize that the expression is now \(\lim_{x \to 0} \left(1 - \frac{1}{2}\right)^{1/x}\), as \(f(x)/x = 1 - \frac{x}{2}\) leads directly to \(-\frac{1}{2}\) when raised to \(1/x\) as \(x\) approaches zero.

Step 5: Apply Limit Properties

By recognizing the form \(\lim_{x \to 0} \left(1 + \frac{-1/2}{x}\right)^{1/x}\), apply the known limit \( \lim_{u \to 0} \left(1 + u\right)^{1/u} = e\) which results in the expression becoming \(e^{-1/2}\).

Step 6: Identify the Correct Answer

The simplified exponential expression is \(e^{-1/2}\). Thus, the correct answer is the option provided in the question, which is (B) \(e^{-1/2}\).

Key Concepts

Taylor Series

The Taylor series is a powerful mathematical tool used to approximate functions. It provides a way to express complex functions as an infinite sum of terms calculated from the values of its derivatives at a single point.
By approximating functions with a polynomial, we can simplify calculations and make complex problems more manageable. This is especially useful near the point where the series is expanded, which is usually zero for simplicity.

• The Taylor series for a function \(f(x)\) around zero is given by \(f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \cdots\).
• For this exercise, we use the Taylor series for \(\cos x\), which is \(\cos x \approx 1 - \frac{x^2}{2} + \frac{x^4}{4!} - \cdots\).

In our problem, only the first few terms are needed to show that \(\cos x \approx 1 - \frac{x^2}{2}\), simplifying our expression when \(x\) is close to zero.

Limit Properties

Limits allow mathematicians to understand the behavior of a function as the input approaches a particular point, in this case zero. Understanding limits is crucial to analyzing and simplifying expressions in calculus, especially when dealing with expressions that become indeterminate, like \(\frac{0}{0}\).
Some helpful properties of limits include:

• \(\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\)
• \(\lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)\)
• If \(f(x)\) is continuous at \(a\), then \(\lim_{x \to a} f(x) = f(a)\)

In solving the limit problem presented, we leveraged the property \(\lim_{u \to 0} \left(1 + u\right)^{1/u} = e\), allowing us to ultimately simplify the expression to \(e^{-1/2}\).

Exponential Form

Exponential forms arise frequently in limits and series due to their unique properties, especially when dealing with the limit definitions like \(\lim_{x \to 0}\). The exponential form \(\left(1 + \frac{f(x)}{x}\right)^{1/x}\) appears often, and using properties of the exponential function can simplify complex limits in calculus.
In the exponential limit equation:\[\lim_{x \to 0} \left(1 + \frac{f(x)}{x}\right)^{1/x} = e^{L}\]
Here, \(L\) is the value that results from the contribution of \(f(x)\) as \(x\) approaches zero. This form is particularly useful because of the approximation \((1 + u)^{1/u} \to e\) as \(u\) approaches zero. In our concept, this allows us to transform a complex-looking limit into a simple expression: \(e^{-1/2}\).
Use of exponential form in calculus facilitates simplification and analysis, showing how intricate expressions contract to a recognizable format.

Mathematical Approximation

Mathematical approximation is an essential aspect of solving calculus problems, as exact solutions are not always feasible due to the complexity of functions. By approximating expressions, we can gain insights into their behavior in simpler terms.
To solve this particular exercise, we approximated \(\cos x\) using a Taylor series to a few terms, recognizing that the contributions of higher degree terms are negligible for small \(x\). This approximation simplified the expression to \(x - \frac{x^2}{2}\), making the limit calculation cleaner.

• Approximations allow for a clearer understanding of problem behavior near specific points, such as zero.
• They reduce the complexity of a problem, which is particularly useful in turning indeterminate forms into solvable ones using familiar limits and exponential form properties.

The solution approach reflects the power of approximation in converting a seemingly complex expression into an elegantly simple exponential form, like \(e^{-1/2}\).