Problem 126 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x}{3 x} \text { \\{Ans. } \frac{2}{3}\right\\} $$

Step-by-Step Solution

Answer

The short answer is: \(\lim _{x \rightarrow 0} \frac{\sin ^{-1} 2 x}{3 x} = \frac{2}{3}\).

Step 1: Identify indeterminate form

As x approaches 0, we can see that the numerator and denominator both approach 0, making this an indeterminate form of type \(0/0\). This suggests we can use L'Hôpital's rule to find the limit.

Step 2: Apply L'Hôpital's rule

L'Hôpital's rule states that if the limit of the ratio of the derivatives of the numerator (f(x)) and denominator (g(x)) exists and is finite, then the limit of the ratio of the original functions also exists and is equal to the limit of the ratio of their derivatives: \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim _{x \rightarrow 0} \frac{f'(x)}{g'(x)}\) provided the limit of the ratio of the derivatives exists. Let's apply L'Hôpital's rule to our function: \(f(x) = sin^{-1}(2x)\), so \(f'(x) = \frac{2}{\sqrt{1 - (2x)^2}}\) \(g(x) = 3x\), so \(g'(x) = 3\) Now, we can rewrite the limit as: \(\lim _{x \rightarrow 0} \frac{\frac{2}{\sqrt{1 - (2x)^2}}}{3}\)

Step 3: Evaluate the limit

Now, we can plug in \(x = 0\) into the limit: \(\lim _{x \rightarrow 0} \frac{2}{\sqrt{1 - (2x)^2}} \div 3 = \frac{2}{\sqrt{1 - (2\cdot0)^2}} \div 3\) Simplifying the expression, we have: \(\frac{2}{\sqrt{1 - 0}} \div 3 = \frac{2}{1} \div 3\) Which gives us the final answer: \(\frac{2}{3}\)

Key Concepts

Indeterminate Form

Indeterminate forms are expressions in calculus that require special methods to evaluate because they appear to be undefined at first glance. One common indeterminate form is \(\frac{0}{0}\). This occurs when both the numerator and the denominator approach zero, which is exactly what happens in this limit problem.
If you encounter an indeterminate form, you can't directly determine the limit using standard calculations. Instead, you need a plan!
For cases like \(\frac{0}{0}\), L'Hôpital's Rule becomes a vital tool. The rule guides you on how to differentiate the numerator and denominator separately and then evaluate the limit of their derivatives.

Inverse Trigonometric Functions

Inverse trigonometric functions "undo" the regular trigonometric functions. They help in finding the angles when you know the trigonometric value. In this exercise, we deal with \(\sin^{-1}(2x)\), also known as the arcsin function.

• The derivative of the arcsin function, \(\sin^{-1}(u)\), with respect to \(u\) is \(\frac{1}{\sqrt{1-u^2}}\).
• Applying the chain rule is necessary when \(u\) itself is not just \(x\), like in \(\sin^{-1}(2x)\). This requires you to multiply by the derivative of the inside function \(2x\), which gives us the full derivative as \(f'(x) = \frac{2}{\sqrt{1-(2x)^2}}\).

Always write carefully to avoid errors, especially when handling functions inside of other functions.

Limit Evaluation

Evaluating limits often involves recognizing patterns or applying special rules like L'Hôpital's Rule. When both parts of a limit approach zero or infinity, like in our problem, straightforward substitution doesn't work.
By following these steps:

• Recognize the indeterminate form, prompting us to use L'Hôpital's.
• Find the derivatives of both the top and bottom functions.
• Evaluate the limit of these derivatives as \(x\) approaches the value in question.

In our solution, we took the derivative of \(\sin^{-1}(2x)\) and \(3x\), then evaluated the limit of their quotient, resulting in a clean \(\frac{2}{3}\).
This approach simplifies complex-looking problems, making challenging calculus questions more manageable. Mastering this method will significantly ease your journey through calculus!"