Problem 265 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \text { \\{Ans. 1\\} } $$

Step-by-Step Solution

Answer

The short answer for the given limit is: \( \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} = 1 \)

Step 1: Identify the indeterminate form

When x approaches 0, the given function approaches an indeterminate form: \( \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} = \frac{0}{0} \)

Step 2: Rationalize the numerator

To eliminate the indeterminate form, we will rationalize the numerator: Multiply the numerator and denominator by the conjugate of the numerator, \( (\sqrt{1+\sin x} + \sqrt{1-\sin x}) \): \( \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} \times \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} \)

Step 3: Simplify the expression

Now, let's simplify the expression: \( \frac{(1+\sin x)-(1-\sin x)}{\tan x(\sqrt{1+\sin x} + \sqrt{1-\sin x})} = \frac{2\sin x}{\tan x(\sqrt{1+\sin x} + \sqrt{1-\sin x})} \) Since \(\tan x = \sin x/\cos x\), we can further simplify the expression: \( \frac{2\sin x}{(\frac{\sin x}{\cos x})(\sqrt{1+\sin x} + \sqrt{1-\sin x})} = \frac{2\cos x}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} \)

Step 4: Find the limit

Now that we've simplified the expression, we can find the limit as x approaches 0 without encountering an indeterminate form: \( \lim _{x \rightarrow 0} \frac{2\cos x}{\sqrt{1+\sin x} + \sqrt{1-\sin x}} = \frac{2\cos (0)}{\sqrt{1+\sin (0)} + \sqrt{1-\sin (0)}} = \frac{2}{\sqrt{1} + \sqrt{1}} = 1 \) So, the required limit is: \( \lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\tan x} = 1 \)

Key Concepts

Indeterminate Forms

When solving limits in calculus, you often encounter expressions where direct substitution leads to an undefined form, like \( \frac{0}{0} \). These are known as **indeterminate forms**. They aren't ready for evaluation without further manipulation. Indeterminate forms can be tricky!

In our problem, when \( x \) approaches 0, both the numerator \( \sqrt{1+\sin x} - \sqrt{1-\sin x} \) and the denominator \( \tan x \) approach 0. That gives us the form \( \frac{0}{0} \). This means it's an indeterminate form, and we have to use additional methods to simplify it into a form where we can determine the limit.

Understanding that an expression is indeterminate is a vital first step. It tells you that plugging in values directly won't work, so another technique like algebraic manipulation or special trigonometric limits is necessary, often making the complex simple.

Rationalization Techniques

Rationalization is a nifty algebraic technique used to simplify expressions involving roots. In this exercise, it helps us remove the indeterminate \( \frac{0}{0} \) form by manipulating the radical expressions.

To rationalize, we multiply the problematic expression by its conjugate. For the given limit problem, the original numerator is \( \sqrt{1+\sin x} - \sqrt{1-\sin x} \). The conjugate of that is \( \sqrt{1+\sin x} + \sqrt{1-\sin x} \). By multiplying the numerator and the denominator by this conjugate:

• Numerator simplifies into a difference of squares: \((1+\sin x) - (1-\sin x) = 2\sin x\).
• The complex radicals are removed, turning the problem into a more manageable form.

After rationalization, you simplify the expression further and are guided to a solution that allows the limit to be evaluated without running into undefined behavior. This gives clarity to the algebraic form and sets things up for eventual cancellation or another algebraic simplification that's solvable.

Trigonometric Limits

Trigonometric limits are crucial when dealing with expressions involving sine, cosine, tangent, and other trigonometric functions. They are the key to solving a lot of calculus problems, especially in limits and derivatives.

In the example, the expression features \( \tan x \), which can be rewritten as \( \frac{\sin x}{\cos x} \). This property makes it easier to simplify the expression after rationalization.

Once simplified, we have \( \frac{2\sin x}{(\frac{\sin x}{\cos x})(\sqrt{1+\sin x} + \sqrt{1-\sin x})} \). From known trigonometric limits, particularly as \( x \to 0 \), \( \sin(0) \) and \( \cos(0) \) meet the conditions for direct substitution, simplifying to 1. With the combination of rationalization techniques and trigonometric identities, you can arrive smoothly at a result.

The takeaway from trigonometric limits is appreciating these functions' limit behaviors, which often simplify otherwise complex settings. Becoming comfortable with standard limits, like \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \), will make calculus much more intuitive and manageable.