Problem 291 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 1} \frac{\sin ^{-1} x}{\tan \frac{\pi x}{2}}\\{\text { Ans. } 0\\} $$

Step-by-Step Solution

Answer

The short answer for the given exercise is that the limit does not exist and the answer is 0.

Step 1: Check if the given limit is indeterminate form

As x approaches 1, the given function becomes: \[ f(x) = \frac{\sin ^{-1}(1)}{\tan \frac{\pi}{2}} \] Since \(\sin^{-1}(1) = \frac{\pi}{2}\) and \(\tan(\frac{\pi}{2})\) is undefined, the given function is in the indeterminate form (\(\frac{0}{0}\)). We need to apply L'Hopital's rule to find the limit.

Step 2: Find the derivative of the numerator and denominator

We need to find the derivatives of both the numerator and denominator. For the numerator, we have: \[ \frac{d}{dx}\left(\sin^{-1}x\right) = \frac{1}{\sqrt{1-x^2}} \] For the denominator, we have: \[ \frac{d}{dx}\left(\tan\frac{\pi x}{2}\right) = \frac{\pi}{2} \cdot \sec^2\frac{\pi x}{2} \]

Step 3: Apply L'Hopital's rule

Now that we have the derivatives for both the numerator and the denominator, we can apply L'Hopital's rule, which states: \[ \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \] Applying L'Hopital's rule to our given limit, we get: \[ \lim_{x \rightarrow 1} \frac{\sin ^{-1} x}{\tan \frac{\pi x}{2}} = \lim_{x \rightarrow 1} \frac{\frac{1}{\sqrt{1-x^2}}}{\frac{\pi}{2} \cdot \sec^2\frac{\pi x}{2}} \]

Step 4: Evaluate the limit after L'Hopital's rule

Now we have to evaluate the limit as x approaches 1: \[ \lim_{x \rightarrow 1} \frac{\frac{1}{\sqrt{1-x^2}}}{\frac{\pi}{2} \cdot \sec^2\frac{\pi x}{2}} = \frac{\frac{1}{\sqrt{1-1^2}}}{\frac{\pi}{2} \cdot \sec^2\frac{\pi \cdot 1}{2}} \] Since \(\sqrt{1-1^2} = 0\) and \(\sec^2(\frac{\pi}{2})\) is undefined, we get the indeterminate form of \(\frac{0}{0}\). We need to apply L'Hopital's rule again.

Step 5: Find the second derivative of the numerator and denominator

To apply L'Hopital's rule a second time, we need to find the second derivatives for both the numerator and the denominator. For the numerator, we have: \[ \frac{d^2}{dx^2}\left(\frac{1}{\sqrt{1-x^2}}\right) = \frac{x}{(1-x^2)^{\frac{3}{2}}} \] For the denominator, we have: \[ \frac{d^2}{dx^2}\left(\frac{\pi}{2} \cdot \sec^2\frac{\pi x}{2}\right) = \frac{\pi^3}{4} \cdot \sec^2\frac{\pi x}{2} \cdot \tan\frac{\pi x}{2} \]

Step 6: Apply L'Hopital's rule a second time

Now that we have the second derivatives for both the numerator and the denominator, we can apply L'Hopital's rule a second time: \[ \lim_{x \rightarrow 1} \frac{\frac{x}{(1-x^2)^{\frac{3}{2}}}}{\frac{\pi^3}{4} \cdot \sec^2\frac{\pi x}{2} \cdot \tan\frac{\pi x}{2}} = \lim_{x \rightarrow 1} \frac{\frac{1}{(1-1^2)^{\frac{3}{2}}}}{\frac{\pi^3}{4} \cdot \sec^2\frac{\pi \cdot 1}{2} \cdot \tan\frac{\pi \cdot 1}{2}} \] Evaluating the limit as x approaches 1, we get: \[ \lim_{x \rightarrow 1} \frac{\sin ^{-1} x}{\tan \frac{\pi x}{2}} = \frac{0}{0} \] This means that the limit does not exist, and the answer to the given exercise is 0.

Key Concepts

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus for dealing with indeterminate forms. These forms often appear when evaluating limits, and they pose challenges due to their undefined nature. L'Hopital's Rule comes into play when direct substitution of a limit results in an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

This rule states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in an indeterminate form, and the derivatives \(f'(x)\) and \(g'(x)\) exist around \(x = a\) (except possibly at \(a\) itself), then:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

It's important to note that L'Hopital's Rule might need to be applied multiple times if the initial application still results in an indeterminate form. Always remember: Use it only if both the numerator and denominator result in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) upon substitution.

Indeterminate Forms

Indeterminate forms occur in calculus when direct evaluation of a limit results in expressions that are undefined. The most common indeterminate forms include \(0/0\) and \(\infty/\infty\). Such forms require further analysis, as the straightforward arithmetic does not apply.

Let's explore why these forms are indeterminate:

• Zero over Zero (\(\frac{0}{0}\)): This can emerge when both the numerator and denominator approach zero. It indicates that more information or a different approach is needed to determine the limit.
• Infinity over Infinity (\(\frac{\infty}{\infty}\)): When both values become enormous, the direct comparison of magnitude isn't straightforward, necessitating a different strategy to evaluate the limit.

Encountering an indeterminate form requires using techniques like L'Hopital's Rule, algebraic manipulation, or series expansion. These approaches help resolve the ambiguity and find the actual limit value.

Derivative Calculations

Calculating derivatives is fundamental in applying L'Hopital's Rule effectively. This involves determining the rate of change of a function and provides critical insight when dealing with indeterminate forms.

For example, in solving the limit given in the exercise, the derivatives of \(\sin^{-1}(x)\) and \(\tan(\frac{\pi x}{2})\) were crucial. Here's how we calculated them:

• The derivative of \(\sin^{-1}(x)\) is: \(\frac{d}{dx}\left(\sin^{-1}x\right) = \frac{1}{\sqrt{1-x^2}}\)
• The derivative of \(\tan(\frac{\pi x}{2})\) involves using the chain rule: \(\frac{d}{dx}\left(\tan\frac{\pi x}{2}\right) = \frac{\pi}{2} \cdot \sec^2\frac{\pi x}{2}\)

Each derivative offers insights into how quickly each function changes as \(x\) approaches a particular value. With these derivatives, we were able to apply L'Hopital's Rule to solve the original limit problem, even repeating the rule to dig deeper when necessary.