Problem 144 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0}(1+\sin x)^{\operatorname{cosecx}}\\{\text { Ans. } e\\} $$

Step-by-Step Solution

Answer

The short version of the answer is: \(\lim_{x \rightarrow 0}(1+\sin x)^{\operatorname{cosecx}} = \boxed{e^{-1}}\).

Step 1: Rewrite the expression

Let's rewrite the given expression as: \[ \lim_{x \rightarrow 0} e^{\ln{(1+\sin x)^{\operatorname{cosecx}}}} \] Now, use the property of logarithms: \[ \lim_{x \rightarrow 0} e^{\operatorname{cosecx} \cdot \ln{(1+\sin x)}} \]

Step 2: Use the limit property

Utilize the property \(\lim_{x \to 0}(1 + f(x))^{\frac{1}{g(x)}} = e^{\lim_{x \to 0}\frac{f(x)}{g(x)}}\): \[ e^{\lim_{x \to 0} \frac{\operatorname{cosecx} \cdot \ln{(1+\sin x)}}{\sin x}} \] In this case, \(f(x) = \operatorname{cosecx} \cdot \ln{(1+\sin x)}\) and \(g(x) = \sin x\).

Step 3: Apply L'Hôpital's Rule

Since the limit of the given function is in indeterminate form, i.e., \(0/0\), we can apply L'Hôpital's Rule to find the limit of \(f(x)/g(x)\): \[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{d(\operatorname{cosecx} \cdot \ln{(1+\sin x)})}{d(\sin x)} \] Compute the derivatives of \(f(x)\) and \(g(x)\): \[ f'(x) = \frac{d(\operatorname{cosecx} \cdot \ln{(1+\sin x)})}{dx} = -\csc x \cot x \cdot \frac{1}{1+\sin x} \cdot \cos x \\ g'(x) = \frac{d(\sin x)}{dx} = \cos x \] Now, compute the limit of the ratio of the derivatives: \[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{-\csc x \cot x \cdot \frac{1}{1+\sin x} \cdot \cos x}{\cos x} \]

Step 4: Compute the limit

Simplify the expression in the limit: \[ \lim_{x \to 0} -\csc x \cot x \cdot \frac{1}{1+\sin x} \] Now, recall that \(\csc x = \frac{1}{\sin x}\) and \(\cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x}\), so we can rewrite the limit as: \[ \lim_{x \to 0} \frac{-\cos x}{1+\sin x} \] Since the limit is no longer in the indeterminate form, it can be directly computed: \[ \lim_{x \to 0} \frac{-\cos x}{1+\sin x} = \frac{-\cos 0}{1 + \sin 0} = \frac{-1}{1} = -1 \]

Step 5: Find the final answer

Now that we have the limit of \(\frac{f'(x)}{g'(x)}\) as \(x \to 0\), we can find the final answer: \[ e^{\lim_{x \to 0} \frac{f'(x)}{g'(x)}} = e^{-1} = \boxed{e^{-1}} \] The final answer to the given limit problem is \(e^{-1}\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter limits that appear indeterminate, you can simplify them using this rule. Here's how it works:

• Check if the limit is an indeterminate form. If it is, you can apply L'Hôpital's Rule.
• Take the derivative of the numerator and the derivative of the denominator separately.
• Re-evaluate the limit using these derivatives.

In our problem, we dealt with an indeterminate \( \frac{0}{0} \) form when evaluating the limit \( \lim_{x \to 0} \frac{f(x)}{g(x)} \). By applying L'Hôpital's Rule, we differentiated the numerator \( f(x) \) and the denominator \( g(x) \), and then found the limit of that new expression to be \(-1\), simplifying the process effectively.

Indeterminate Forms

Indeterminate forms appear in calculus when the limit cannot be directly evaluated. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and more.

Recognizing these forms is crucial because they indicate that a straightforward evaluation will not work. In our exercise, the initial form \( \lim_{x \to 0} \frac{\operatorname{cosecx} \cdot \ln{(1+\sin x)}}{\sin x} \) becomes \( \frac{0}{0} \) as \( x \to 0 \).

• This is because \( \sin x \to 0 \), making both numerator and denominator zero.
• Indeterminate forms require further steps to resolve, often using methods like L'Hôpital's Rule or algebraic manipulation.

Being able to identify these forms helps you decide on the correct method, ensuring you don’t get stuck in complicated calculations.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent play a crucial role in calculus problems. They describe the relationships of angles in right triangles, circular motion, and oscillations.

In the context of limits, we often work with functions like \( \sin x \), \( \cos x \), \( \tan x \), \( \csc x \) (cosecant), and \( \cot x \) (cotangent):

• \( \sin x \) and \( \cos x \) are periodic functions that oscillate between -1 and 1.
• \( \csc x \) is the reciprocal of \( \sin x \), and \( \cot x \) is the reciprocal of \( \tan x \).

In the solution, recognizing how \( \csc x = \frac{1}{\sin x} \) and \( \cot x = \frac{\cos x}{\sin x} \) help simplified the terms in the expression.
Understanding these functions not only aids in solving limits but also enriches comprehension of wave phenomena and other periodic behaviors.