Problem 19 - A Textbook of Differential Calculus

Question

If \(\lim _{x \rightarrow \infty} 4 x\left(\frac{\pi}{4}-\tan ^{-1} \frac{x+1}{x+2}\right)=y^{2}+4 y+5\), then \(y\) can be equal to (a) 1 (b) \(-1\) (c) \(-4\) (d) \(-3\)

Step-by-Step Solution

Answer

No real value of \(y\). None of the options are correct.

Step 1: Evaluate the Limit Expression

We are given the limit expression: \( \lim _{x \rightarrow \infty} 4x \left( \frac{\pi}{4} - \tan^{-1}\left( \frac{x+1}{x+2} \right) \right) \). As \( x \to \infty \), \( \frac{x+1}{x+2} \approx 1 \). Thus, \( \tan^{-1}\left( \frac{x+1}{x+2} \right) \to \tan^{-1}(1) = \frac{\pi}{4} \). So, \( \frac{\pi}{4} - \tan^{-1}\left( \frac{x+1}{x+2} \right) \approx 0 \).

Step 2: Apply L'Hopital's Rule

Since both numerator and denominator tend to 0 as \(x \to \infty\), we apply L'Hopital's Rule. If we set \( f(x) = \frac{\pi}{4} - \tan^{-1}\left( \frac{x+1}{x+2} \right) \), the expression becomes: \( \lim _{x \rightarrow \infty} \frac{\frac{\pi}{4} - \tan^{-1}\left( \frac{x+1}{x+2} \right)}{\frac{1}{4x}} \).Differentiating the numerator and denominator with respect to \(x\), we get:- Derivative of numerator \(f'(x) = \frac{1}{2 \cdot (x+2)^2} \) (chain rule and derivative of \(\tan^{-1}(x)\) form).- The derivative of the denominator w.r.t. \(x\) is \(- \frac{1}{4x^2}\).Thus, the limit simplifies to: \( \lim _{x \rightarrow \infty} \frac{\frac{1}{2(x+2)^2}}{-\frac{1}{4x^2}} = \lim _{x \rightarrow \infty} \frac{-2x^2}{(x+2)^2} \).

Step 3: Simplify Resulting Limit Expression

We now simplify the expression from the previous step: \(-2 \cdot \left( \frac{x^2}{(x+2)^2} \right)\). As \(x \to \infty\), \(\frac{x^2}{(x+2)^2} \approx 1\), because the effect of \(+2\) becomes negligible. Thus, the limit becomes \(-2\).

Step 4: Equate Limit Result to Quadratic Expression

The result of the limit is \( -2 \). We know \(-2 = y^2 + 4y + 5\). This is a quadratic equation in\(y\): \( y^2 + 4y + 7 = 0\).

Step 5: Solve the Quadratic Equation

To solve the equation \( y^2 + 4y + 7 = 0 \), use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 4, c = 7 \).Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \times 1 \times 7 = 16 - 28 = -12 \]Since the discriminant is negative, there are no real solutions.

Key Concepts

L'Hopital's Rule

Understanding L'Hopital's Rule in calculus can be very helpful when evaluating limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule allows us to differentiate the numerator and the denominator separately, simplifying the process of finding the limit. Essentially, it states that if the limit of \( \frac{f(x)}{g(x)} \) exists as \( x \to c \), and both \( f(x) \) and \( g(x) \) approach zero (or both approach infinity), you can compute:

• The limit as \( \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \), provided this limit exists.

In our exercise, the expression is rewritten using L'Hopital's Rule to deal with the indeterminate form by taking derivatives of the numerator and denominator. Differentiating correctly is crucial here. For example, the derivative of \( \tan^{-1}\left(\frac{x+1}{x+2}\right) \) requires applying the chain rule well. Overall, L'Hopital's Rule simplifies complex limit problems into more manageable computations.

Quadratic Equations

Quadratic equations often arise in mathematics and can be recognized by their standard form \( ax^2 + bx + c = 0 \). Solving these can uncover important insights about the problem at hand. In some cases, the solutions can be easily found using factoring, but when factoring is difficult, the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is very useful.

In this exercise, we reach a scenario where we have to solve a quadratic equation \( y^2 + 4y + 7 = 0 \). By calculating the discriminant \( b^2 - 4ac \), we quickly determine the nature of its roots. Here, since the discriminant \( (-12) \) is negative, it tells us that there are no real solutions. In mathematical terms, this indicates the lack of intersections of the quadratic curve with the x-axis, meaning no real values of \( y \) satisfy this quadratic equation.

Trigonometric Limits

Trigonometric limits are a foundational concept in calculus, often involving functions like sine, cosine, and tangent. These functions become particularly interesting when their arguments approach critical points, such as zero or infinity, as they can lead to indeterminate forms demanding careful analysis.

• By understanding basic trigonometric identities and their limit properties, you can simplify complex expressions effectively.

In the provided problem, one component involves the inverse tangent function, \( \tan^{-1}(x) \), which is known for approaching \( \frac{\pi}{4} \) as its input approaches 1. This behavior is crucial to grasp, helping us understand the results as \( x \to \infty \). Recognizing such symmetrical behavior is helpful in predicting outcomes of limits that initially seem complex. When thoroughly understood, trigonometric limits provide elegant tools to simplify and solve a wide array of calculus problems.