Problem 382 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \frac{\pi}{2}}(\sin x)^{\tan x} \quad\\{\text { Ans. } 1\\} $$

Step-by-Step Solution

Answer

The limit of the given function as x approaches \(\pi/2\) is \(1/e^2\).

Step 1: Substitute

Substitute \(x = \pi/2\). After substitution, the function becomes \((\sin(\pi/2))^{\tan(\pi/2)}\). As \(\sin(\pi/2)\) equals 1 and \(\tan(\pi/2)\) is undefined, the function equates to \(1^{\infty}\) which is an indeterminate form.

Step 2: Express in exponential form

When dealing with the '1 to the power infinity' indeterminate form, we can convert the function into an exponential form : \(e^{f(x)*g(x)}\). In this case, we write it as \( e^{\lim_{x\rightarrow\frac{\pi}{2}}[\tan(x)(\ln(\sin(x)))]}\).

Step 3: Apply L'Hopital's Rule

The current limit is also in an indeterminate form of '0 times infinity'. We can rewrite it as \(\frac{\ln(\sin(x))}{\cot(x)}\) ? trying to convert it into '0/0' or 'infinity/infinity', and use L'Hopital's Rule. Before applying the rule, let's simplify more as follows : \(e^{\lim_{x\rightarrow\frac{\pi}{2}}\left(\frac{\ln(\sin(x))}{\cot(x)}\right)}\) = \(e^{\lim_{x\rightarrow\frac{\pi}{2}}\left(\frac{\ln(1-\cos^2(x))}{\frac{\cos(x)}{\sin(x)}}\right)}\). Now, apply L'Hopital's rule gives : \(e^{\lim_{x\rightarrow\frac{\pi}{2}}\left(\frac{-2\cos(x)*\sin(x)}{\cos^2(x)+\sin^2(x)}\right)}\).

Step 4: Solving the limit

Substituting \(x = \pi/2\) now gives : \(e^{\lim_{x\rightarrow\frac{\pi}{2}}(-2)} = 1/e^2\).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a valuable tool for finding limits of indeterminate forms. If you have a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), L'Hôpital's Rule allows you to differentiate the numerator and denominator separately. This can often simplify the expression and make the limit easier to solve.
In our exercise, we faced an indeterminate form while evaluating \( \frac{\ln(\sin(x))}{\cot(x)} \). After rewriting it into a fraction, applying L'Hôpital's Rule leads to finding the derivatives of \( \ln(\sin(x)) \) and \( \cot(x) \).

• Differentiate \( \ln(\sin(x)) \) to get \( \frac{1}{\sin(x)} \cdot \cos(x) \)
• Differentiate \( \cot(x) \) to get \(-\csc^2(x) \)

By substituting these derivatives back into the limit, you simplify and solve the original problem. This technique helps us avoid the complications of directly dealing with undefined forms.

Indeterminate Forms

Indeterminate forms appear when a limit leads to expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or \( 1^{\infty} \). These forms don't immediately reveal the limit’s value, so special methods are used to resolve them.
In the given problem, substituting \( x = \frac{\pi}{2} \) resulted in \( (\sin(\pi/2))^{\tan(\pi/2)} \), turning into \( 1^{\infty} \), an indeterminate form. To manage this, we used exponential form rewriting: \( e^{f(x) \cdot g(x)} \). This steps around the immediate "undefined" nature and provides a route to evaluate the limit.
Using methods like rewriting functions into exponential or applying L'Hôpital’s rule, you navigate indeterminate forms and simplify the evaluation process.

Exponential Functions

Exponential functions take the form \( e^{f(x)} \), where \( e \) is the base of natural logarithms. They are crucial in calculus, particularly when dealing with indeterminate forms.
When encountering \( 1^{\infty} \), as we saw in the example, rewriting the expression as \( e^{f(x) \cdot g(x)} \) transforms the problem. This makes it manageable under the rules of limits and differentiation.
In our exercise, converting \( (\sin(x))^{\tan(x)} \) into an exponential function: \( e^{\tan(x) \cdot \ln(\sin(x))} \), facilitated using calculus to evaluate its limit.

• This conversion aligns with logarithmic properties, simplifying the handling of exponentiated forms.
• Allows application of L'Hôpital’s rule to resolve indeterminate scenarios.

Recognizing when exponential transformation is beneficial is key to dealing with complex limit problems effectively.