Problem 76 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The value is \( e^{-1/2} \), corresponding to option (B).

Step 1: Identify the limit form

The expression given is \(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{1 / x}\). We need to determine if it matches any well-known limit forms.

Step 2: Recognize the indeterminate form

First, evaluate the expression \(\frac{x - 1 + \cos x}{x}\) as \(x \to 0\). Substituting \(x = 0\), we get \(\frac{0 - 1 + 1}{0} = \frac{0}{0}\), which is an indeterminate form. Additionally, the whole expression is of the form \(1^{\infty}\) which is a standard indeterminate form for limits.

Step 3: Apply logarithms to simplify

To handle the \(1^{\infty}\) form, take the natural logarithm of the expression: \(y = \left(\frac{x - 1 + \cos x}{x}\right)^{1/x}\). This becomes \(\ln y = \frac{1}{x} \ln \left(\frac{x - 1 + \cos x}{x}\right)\).

Step 4: Simplify the logarithm

Consider \(\ln \left(\frac{x - 1 + \cos x}{x}\right) = \ln(x - 1 + \cos x) - \ln x\). This requires a further expansion using L'Hopital's Rule since both \(x - 1 + \cos x\) and \(x\) tend to 0 as \(x \to 0\).

Step 5: Apply L'Hopital's Rule

Define \(f(x) = \ln(x - 1 + \cos x)\) and \(g(x) = \ln x\). Apply L'Hopital's Rule to \(\lim_{x \to 0} \frac{\ln(x - 1 + \cos x) - \ln x}{x}\). Differentiate both numerators and denominators separately and evaluate the limit.

Step 6: Calculate derivatives

The derivative of \(f(x) = \ln(x - 1 + \cos x)\) is \(f'(x) = \frac{1}{x - 1 + \cos x}(1 - \sin x)\). The derivative of \(g(x) = \ln x\) is \(g'(x) = \frac{1}{x}\).

Step 7: Evaluate the limit

Calculate the limit \(\lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \left(\frac{x(1 - \sin x)}{x - 1 + \cos x}\right).\) Substitute \(x = 0\) to get \(-\frac{1}{2}\).

Step 8: Exponential form

Therefore, \(\ln y = -\frac{1}{2}\) as \(x \to 0\). Hence \(y = e^{-\frac{1}{2}}\). The solution is \(y = \lim_{x \to 0}\left(\frac{x-1+\cos x}{x}\right)^{1/x} = e^{-1/2}.\)

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms appear when substituting a limit into a function results in an ambiguous expression. For instance, the expression \(0/0\) can suggest infinitely many outcomes, and thus, analysis is required to determine the actual limit. Indeterminate forms come in a variety of types, including:

• \(0/0\)
• \(\infty/\infty\)
• \(0 \cdot \infty\)
• \(1^\infty\)
• \(\infty - \infty\)
• \(0^0\)
• \(\infty^0\)

When faced with these forms, techniques like simplification, factorization, or L'Hopital's Rule can be employed to find the limit. In our exercise, the limit of the form \(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{1 / x}\) translates to \(1^\infty\), indicating further analysis is required to unlock the solution.

L'Hopital's Rule

L'Hopital's Rule is a valuable tool in calculus for resolving limits that present indeterminate forms like \(0/0\) or \(\infty/\infty\). The rule states that if a limit is presented as \(\lim_{x \to c} \frac{f(x)}{g(x)}\) and evaluates to either \(0/0\) or \(\infty/\infty\), then the limit can be resolved using derivatives: \[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]provided the latter limit exists. In practice, this method involves:

• Identifying the indeterminate form
• Differentiating the numerator \(f(x)\) and the denominator \(g(x)\) separately
• Evaluating the new limit with these derivatives

In our problem, part of the expression \(\ln \left(\frac{x - 1 + \cos x}{x}\right)\) is resolved with L'Hopital's Rule after identifying that both numerator and denominator approach zero as \(x \to 0\). This enables us to calculate the limit correctly through differentiation.

Natural Logarithm

The natural logarithm, expressed as \(\ln(x)\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. Natural logarithms have unique properties that simplify complex expressions and make them very useful in calculus, particularly in limit problems. Here are some key properties of natural logarithms:

• \(\ln(ab) = \ln a + \ln b\)
• \(\ln\left(\frac{a}{b}\right) = \ln a - \ln b\)
• \(\ln(a^b) = b \ln a\)

In solving limits, especially those involving exponentials or power expressions, taking the natural logarithm of an expression can linearize it, thereby making it easier to handle. In our solution, the initial expression in the exercise is of \(1^\infty\) form. Applying natural logarithms transforms the limit into a linear expression, facilitating easy differentiation and application of calculus techniques such as L’Hopital's Rule.