Problem 53 - Single Variable Calculus

Question

Evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow \infty} \frac{2 \tan ^{-1} x-\pi}{e^{1 / x^{2}}-1} $$

Step-by-Step Solution

Answer

The limit of the given expression as \(x\rightarrow\infty\) is 0, found using L'Hôpital's Rule.

Step 1: Applying L'Hôpital's Rule

Since both the numerator and the denominator go to infinity as \(x \rightarrow \infty\), we can apply L'Hôpital's Rule by finding the derivatives of both the numerator and the denominator with respect to x. Let's find the derivative of \(2 \tan ^{-1} x\): \[ \frac{d}{dx} (2 \tan^{-1} x) = 2 \frac{1}{1 + x^2} \] Let's find the derivative of \(e^{1/x^2} - 1\): \[ \frac{d}{dx} (e^{1/x^2} - 1) = -\frac{2}{x^3} e^{1/x^2} \] Now that we have the derivatives, let's take the limit.

Step 2: Calculating the limit of derivatives

Apply L'Hôpital's Rule by taking the limit of the derivatives as \(x \rightarrow \infty\): \[ \lim _{x \rightarrow \infty} \frac{2 \tan ^{-1} x-\pi}{e^{1 / x^{2}}-1} = \lim _{x \rightarrow \infty} \frac{2 \frac{1}{1 + x^2}}{-\frac{2}{x^3} e^{1/x^2}} \] Now, we can simplify the expression:

Step 3: Simplifying the expression

Multiply the second term in the numerator by \(x^3\) and the second term in the denominator by \(x^3\) to get: \[ \lim _{x \rightarrow \infty} \frac{2x^3 \frac{1}{1 + x^2}}{-2 e^{1/x^2}} \] Cancelling out the 2's: \[ \lim _{x \rightarrow \infty} \frac{x^3 \frac{1}{1+x^2}}{-( e^{1/x^2})}. \] Now, we can find the limit by computing the limit of the numerator (which is zero) and the limit of the denominator (which is negative 1). \[ \lim_{x\to\infty}\frac{x^3\frac{1}{1+x^2}}{-e^{1/{x^2}}} = \frac{0}{-1} = 0. \] Thus, the limit of the given expression as \(x\rightarrow\infty\) is 0, using L'Hôpital's Rule where appropriate.

Key Concepts

Limits at Infinity

Understanding limits at infinity is a fundamental concept in calculus, particularly when we assess the behavior of functions as they grow without bound. When determining a limit at infinity, one aims to find the value a function approaches as the variable either increases or decreases without limit.

For instance, when evaluating the limit of a ratio of functions as in the exercise, if both the numerator and denominator grow infinitely large, the limit isn't immediately obvious. This is where L'Hôpital's Rule comes into play, which states that if the limits of both the numerator and the denominator are zero or both are infinity, the limit of the ratio can be found by taking the derivatives of the numerator and denominator. Applying this rule often simplifies the expression and makes the limit easier to identify.

Derivative of Inverse Trigonometric Functions

When working with inverse trigonometric functions, like arctangent, the derivative provides the rate of change of the function. For example, the derivative of arctangent of x, denoted as \(\tan^{-1}x\), is \(\dfrac{1}{1 + x^2}\).

In the given exercise, the derivative of \(2\tan^{-1}x\) is necessary for applying L'Hôpital's Rule. We found that by using the chain rule and understanding that \(\tan^{-1}x\) is the inverse of the tangent function which provides us with a slope of the tangent line for the angle whose tangent value is x. The procedure involves differentiating with respect to x to simplify complex limits involving these functions, as demonstrated in the step-by-step solution.

Exponential Functions

The behavior of exponential functions can often be counterintuitive, especially as variables approach infinity. For the function \(e^x\), where e is the base of the natural logarithm, the function grows rapidly as x increases. The derivative provides us with a powerful tool for understanding how the function changes at any given point.

In our exercise solution, the derivative of the function \(e^{1/x^2} - 1\) is taken. The power rule and chain rule for derivatives are used to find the rate of change of the exponential function with respect to x. Exponential functions are unique in that the rate of change at any given point is directly proportional to the value of the function at that point, which directly affects how we understand their limits at infinity.