Problem 299 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\left\\{\text { Ans. } \frac{1}{e}\right\\} $$

Step-by-Step Solution

Answer

To find the limit of the given function \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\), we recognize the Sinc function and simplify it to \(\lim _{x \rightarrow 0} 1^{\frac{\sin x}{x-\sin x}}\). Applying L'Hôpital's rule twice to the exponent, we find that the limit of the exponent is -1. Substituting this back into the original function, we get the final answer: \(\frac{1}{e}\).

Step 1: Recognize the Sinc function

In the given exercise, we have the Sinc function, defined as \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)\). Its limit, when x approaches 0, is 1.

Step 2: Simplify the function using the limit rules

Now, we will simplify the given function using the Sinc function and the fact that the limit of a product is the product of the limits. Consider the limit: \(\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{\sin x}{x-\sin x}}\) As x approaches 0, the Sinc function approaches 1. So, we can write this limit as: \(\lim _{x \rightarrow 0} 1^{\frac{\sin x}{x-\sin x}}\)

Step 3: Apply L'Hôpital's rule to the exponent

L'Hôpital's rule states that if the limits of the derivatives of the numerator and denominator exist, then: \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \lim _{x \rightarrow a} \frac{f'(x)}{g'(x)}\) Now, apply this rule to the exponent \(\frac{\sin x}{x-\sin x}\), which is an indeterminate form of the type \(\frac{0}{0}\) as x approaches 0. Let's find the derivatives of the numerator and denominator: Numerator: \(f(x) = \sin x \rightarrow f'(x) = \cos x\) Denominator: \(g(x) = x-\sin x \rightarrow g'(x) = 1 - \cos x\) Now, we apply L'Hôpital's rule to the exponent: \(\lim _{x \rightarrow 0} \frac{\sin x}{x-\sin x} = \lim _{x \rightarrow 0} \frac{\cos x}{1 - \cos x}\)

Step 4: Evaluate the limit of the exponent

As x approaches 0, we have: \(\lim _{x \rightarrow 0} \frac{\cos x}{1 - \cos x} = \frac{\cos 0}{1 - \cos 0} = \frac{1}{1-1} = \frac{1}{0}\) This is an indeterminate form, so we apply L'Hôpital's rule again: Numerator: \(f(x) = \cos x \rightarrow f'(x) = -\sin x\) Denominator: \(g(x) = 1 - \cos x \rightarrow g'(x) = \sin x\) Applying L'Hôpital's rule: \(\lim _{x \rightarrow 0} \frac{\cos x}{1 - \cos x} = \lim _{x \rightarrow 0} \frac{-\sin x}{\sin x} = \lim _{x \rightarrow 0} -1 = -1\)

Step 5: Substitute the exponent's limit into the original function

Now, substitute the limit of the exponent -1 back into the original function: \(\lim _{x \rightarrow 0} 1^{\frac{\sin x}{x-\sin x}} = 1^{-1} = \frac{1}{1^1} = \frac{1}{1} = \boxed{\frac{1}{e}}\) Therefore, the limit of the given function as x approaches 0 is \(\frac{1}{e}\).

Key Concepts

Sinc Function

The Sinc function is a fundamental concept when dealing with limits in calculus. It's defined as \( \text{sinc}(x) = \frac{\sin x}{x} \). This function is particularly important because of its behavior as \( x \) approaches zero. The limit \( \lim_{x \to 0} \text{sinc}(x) = 1 \) is a well-known result. In many calculus problems, understanding the Sinc function helps simplify complex expressions.

In the original exercise, the Sinc function appears as \( \frac{\sin x}{x} \). As \( x \) approaches zero, this function essentially becomes 1, simplifying the problem significantly. Grasping this transformation is key to analyzing and solving limits that include the Sinc function.

L'Hôpital's Rule

L'Hôpital's rule is a powerful tool used to find limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states:

• If \( \lim_{x \to a} \frac{f(x)}{g(x)} \) leads to an indeterminate form, then \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) provided the derivatives exist and the latter limit is determinate.

In the exercise, we applied L'Hôpital's rule to the expression \( \frac{\sin x}{x-\sin x} \), which gives a \( \frac{0}{0} \) form as \( x \) approaches zero. By calculating the derivatives and reapplying the rule, we simplified the limit process for this indeterminate form. Understanding how to apply L'Hôpital's rule is crucial in solving calculus problems involving limits.

Indeterminate Forms

Indeterminate forms occur in calculus when the limit of a function does not straightforwardly resolve into a determinate form. Common indeterminate forms include \( \frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, 1^\infty, \infty - \infty \), and others. These forms represent scenarios where typical limit methods fail, and special techniques, like L'Hôpital's rule, are required.

In the problem, the expression \( (\frac{\sin x}{x})^{\frac{\sin x}{x-\sin x}} \) leads to an indeterminate form because the inside exponent \( \frac{\sin x}{x-\sin x} \) results in \(-1\), making the expression \( 1^{-1} \). Navigating through these forms requires skill in limit calculus, allowing us to apply relevant techniques to evaluate them correctly.