Problem 361 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\tan ^{-1} 3 x}-\sqrt[3]{1-\sin ^{-1} 3 x}}{\sqrt{1-\sin ^{-1} 2 x}-\sqrt{1+\tan ^{-1} 2 x}} \text { \\{Ans. -1\\} } $$

Step-by-Step Solution

Answer

We are given the limit \(\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+\tan ^{-1} 3 x}-\sqrt[3]{1-\sin ^{-1} 3 x}}{\sqrt{1-\sin ^{-1} 2 x}-\sqrt{1+\tan ^{-1} 2 x}}\). Evaluating the numerator and denominator individually at x = 0 results in an indeterminate form \(\frac{0}{0}\). To find the limit, we apply L'Hôpital's Rule and compute the derivatives of the numerator and denominator. After plugging x = 0 into the limit of the ratio of derivatives, we find the answer to be \(\boxed{-1}\).

Step 1: Evaluate the limits of the numerator and the denominator individually

: First, attempt to evaluate the limits of the numerator and the denominator separately by substituting x=0 and see if any indeterminate form or issues arise. For the numerator, we have: \[\lim_{x\to 0}\left(\sqrt[3]{1+\tan^{-1}{3x}}-\sqrt[3]{1-\sin^{-1}{3x}}\right)\] As x goes to 0, both \(\tan^{-1}{3x}\) and \(\sin^{-1}{3x}\) go to 0 as well. Therefore, the limit of the numerator becomes: \[\lim_{x\to 0}\left(\sqrt[3]{1}-\sqrt[3]{1}\right) = 0\] Now, evaluate the limit of the denominator: \[\lim_{x\to 0}\left(\sqrt{1-\sin^{-1}{2x}}-\sqrt{1+\tan^{-1}{2x}}\right)\] Again, as x approaches 0, both \(\tan^{-1}{2x}\) and \(\sin^{-1}{2x}\) go to 0 as well. Therefore, the limit of the denominator becomes: \[\lim_{x\to 0}\left(\sqrt{1}-\sqrt{1}\right) = 0\] Now we have the indeterminate form, \(\frac{0}{0}\), so we need to find another way to evaluate the limit.

Step 2: Apply L'Hôpital's Rule

: To tackle the issue of indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit of the ratio of the derivatives of the functions exists, then the limit of the original functions also exists and is equal to the limit of the ratio of derivatives. We will first need to find the derivatives of the functions in the numerator and denominator. 1. Numerator: \[\frac{d}{dx}\left(\sqrt[3]{1+\tan^{-1}{3x}}-\sqrt[3]{1-\sin^{-1}{3x}}\right)\] 2. Denominator: \[\frac{d}{dx}\left(\sqrt{1-\sin^{-1}{2x}}-\sqrt{1+\tan^{-1}{2x}}\right)\] Now, calculate the derivatives: 1. Numerator: \[\frac{1}{3\sqrt[3]{(1+\tan^{-1}{3x})^2}}\cdot\frac{3}{1+(3x)^2}\cdot3 - \frac{1}{3\sqrt[3]{(1-\sin^{-1}{3x})^2}}\cdot\frac{3}{\sqrt{1-(3x)^2}}\cdot (-3)\] 2. Denominator: \[\frac{-1}{2\sqrt{1-\sin^{-1}{2x}}}\cdot\frac{2}{\sqrt{1-(2x)^2}}\cdot (-2) - \frac{1}{2\sqrt{1+\tan^{-1}{2x}}}\cdot\frac{2}{1+(2x)^2}\cdot 2\] Now apply L'Hôpital's Rule: \[\lim _{x \rightarrow 0} \frac{\frac{1}{3\sqrt[3]{(1+\tan^{-1}{3x})^2}}\cdot\frac{3}{1+(3x)^2}\cdot3 - \frac{1}{3\sqrt[3]{(1-\sin^{-1}{3x})^2}}\cdot\frac{3}{\sqrt{1-(3x)^2}}\cdot (-3)}{\frac{-1}{2\sqrt{1-\sin^{-1}{2x}}}\cdot\frac{2}{\sqrt{1-(2x)^2}}\cdot (-2) - \frac{1}{2\sqrt{1+\tan^{-1}{2x}}}\cdot\frac{2}{1+(2x)^2}\cdot 2}\] As x approaches 0, we have: \[\frac{\frac{1}{3\cdot 1}\cdot\frac{3}{1} - \frac{1}{3\cdot 1}\cdot\frac{3}{1}}{\frac{-1}{2\cdot 1}\cdot\frac{2}{1} - \frac{1}{2\cdot 1}\cdot\frac{2}{1}} = \frac{1 - (-1)}{-1 - 1} =\] \[-1\] Hence, the limit of the given expression is -1.

Key Concepts

L'Hôpital's Rule

Understanding the application of L'Hôpital's Rule can be crucial when determining seemingly unsolvable limits. This mathematical principle provides us with a solution whenever we encounter an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The rule states that under certain conditions, the limit of a ratio of two functions can be found by taking the limit of the ratio of their derivatives.

In the exercise presented, the initial attempt to calculate the limit results in an indeterminate form. By applying L'Hôpital's Rule, we proceed to differentiate both the numerator and the denominator independently. After differentiating, we substitute the approaching value into the derivatives and simplify the expression. The key to using L'Hôpital's Rule effectively lies in correctly finding the derivatives and ensuring that the ratio of the derivatives does not yield another indeterminate form. If that's the case, L'Hôpital's Rule may be applied repeatedly until a determinate form is achieved.

Indeterminate Form

In calculus, when we speak of an indeterminate form, we're referring to a mathematical expression that does not have a well-defined limit as it stands. The most common types are \(0/0\), \(\infty/\infty\), \(0\cdot\infty\), \(\infty - \infty\), \(0^0\), \(\infty^0\), and \(1^\infty\). They arise when the limits we are evaluating are confronted with operations that do not produce a single, unique result.

In the given exercise, by directly substituting \(x\) with 0, we obtained the indeterminate form \(0/0\). Without additional methods such as L'Hôpital's Rule, it would be impossible to determine the exact value of the limit. These forms indicate that further analysis is needed to solve the limit and that simply substituting the variable is not enough.

Trigonometric Limits

Trigonometric limits are a key part of calculus, particularly when the functions involved include trigonometric expressions such as \(\sin\), \(\cos\), \(\tan\), and their inverses. These limits can often lead to indeterminate forms, especially when the variable is approaching a critical point where the function is undefined or where its behavior changes drastically.

In the exercise, we see inverse trigonometric functions, like \(\tan^{-1}(3x)\) and \(\sin^{-1}(2x)\), both of which contribute to the indeterminacy when \(x\) approaches 0. The solution to these tricky limits often involves trigonometric identities or, as illustrated in the exercise, advanced calculus methods like L'Hôpital's Rule. Understanding the behavior of trigonometric functions and their limits near critical points is crucial for studying continuous and differentiable behavior of functions in calculus.