Problem 307 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \infty} \frac{e^{\frac{1}{x^{2}}}-1}{2 \tan ^{-1} x^{2}-\pi}\left\\{\text { Ans. }-\frac{1}{2}\right\\} $$

Step-by-Step Solution

Answer

Based on the given solution, the short answer is that the limit of the given function, as x approaches infinity, is \(-\frac{1}{2}\). This is determined using L'Hôpital's Rule and finding the derivatives of both the numerator and the denominator, then calculating their limit.

Step 1: Verify that L'Hôpital's Rule can be applied

First, we need to check if we can use L'Hôpital's Rule to find the given limit. Let's plug in x=∞ into the function: \[ \frac{e^{\frac{1}{\infty^2}} - 1}{2 \tan^{-1}(\infty^2) - \pi} \] As x approaches infinity, \(\frac{1}{x^2}\) approaches 0. Therefore, the numerator approaches \(e^0 - 1 = 0\). The arctangent of \(x^2\) approaches \(\frac{\pi}{2}\) and the denominator approaches \(2 \cdot \frac{\pi}{2} - \pi = 0\). Since both the numerator and denominator approach 0, we can use L'Hôpital's Rule.

Step 2: Find the derivatives of the numerator and the denominator

Now, we need to find the derivatives of the numerator and the denominator with respect to x. Let \(f(x) = e^{\frac{1}{x^2}} - 1\) and \(g(x) = 2 \tan^{-1}(x^2) - \pi\). The derivative of \(f(x)\) with respect to x is: \[ f'(x) = -2x \cdot e^{\frac{1}{x^2}} \] The derivative of \(g(x)\) with respect to x is: \[ g'(x) = 4x \cdot \frac{1}{1 + (x^2)^2} \]

Step 3: Apply L'Hôpital's Rule

Now, we will apply L'Hôpital's Rule to find the limit: \[ \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow \infty} \frac{-2x \cdot e^{\frac{1}{x^2}}}{4x \cdot \frac{1}{1 + (x^2)^2}} \] Now, we can simplify the expression a little: \[ = \lim_{x \rightarrow \infty} \frac{-2e^{\frac{1}{x^2}}}{4 \cdot \frac{1}{1 + x^4}} \] Now, as x approaches infinity, the numerator approaches -2 and the denominator approaches 0 from the right-hand side as \(x^4\) becomes very large. This makes the whole fraction approach \(-\frac{1}{2}\). Hence, the limit of the given function is \(-\frac{1}{2}\).

Key Concepts

Limits at Infinity

Understanding limits at infinity is essential for solving complex calculus problems—particularly when assessing the behavior of a function as the variable grows without bound.

When we talk about the limit of a function as x approaches infinity, we're describing what value the function is getting closer to as the value of x increases towards infinity. This can often result in a finite value, infinity, negative infinity, or it may not exist at all.

In our exercise, the limit approaches a form known as an indeterminate form, specifically 0/0. This is where L'Hôpital's Rule shines, helping us determine the true limit by comparing the rates of change of the numerator and denominator—essentially a race between how fast the top and bottom of our fraction are heading towards zero. With appropriate derivatives in hand, we can often resolve these indeterminate forms and find a clear limit.

Derivatives in Calculus

The derivative represents the rate at which a function is changing at any given point and is foundational in calculus. It's the tool we use to calculate instantaneous rates of change and slopes of tangents to curves.

Within the solution for our exercise, derivatives allowed us to apply L'Hôpital's Rule effectively. The steps show how to calculate the derivative of both the numerator and the denominator (f'(x) and g'(x)). We need derivatives when the limit of quotient of functions results in an indeterminate form, like 0/0 or ∞/∞. By taking derivatives and simplifying, we're able to eliminate the indeterminate form and find what value the expression is actually tending towards as x tends to infinity. It's important to remember that the derivative is a limit itself, a concept that connects the central ideas of calculus in a beautiful symmetry.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent—and their inverses—have profound applications in calculus, especially when dealing with limits and derivatives involving angles.

For instance, in our problem we are dealing with the arctangent, which is the inverse of the tangent function. As the variable within the arctangent function grows, the output of the arctangent function approaches \(\frac{\pi}{2}\), which is the limit of the arctangent as x approaches infinity. This behavior of trigonometric functions at infinity is key to understanding how to deal with complex limits and is particularly handy in solving trigonometric limits using L'Hôpital's Rule. The derivatives of these functions often involve the other trigonometric functions, as demonstrated in the solution, and understanding their relationships is crucial for solving calculus problems.