Problem 32 - A Textbook of Differential Calculus

Question

\(\operatorname{Let} f(x)=\lim _{n \rightarrow-}\left(\cos \sqrt{\frac{x}{n}}\right)^{n}, g(x)=\lim _{n \rightarrow-}(1+x+x \sqrt[n]{e})^{n} .\) Now, consider the function \(y=h(x)\), where \(h(x)=\tan ^{-1}\left(g^{-1} f^{-1}(x)\right)\) \(\lim _{x \rightarrow 0} \frac{\ln (f(x))}{\ln (g(x))}\) is equal to (a) \(\frac{1}{2}\) (b) \(-\frac{1}{2}\) (c) 0 (d) 1

Step-by-Step Solution

Answer

The answer is (b) \(-\frac{1}{2}\).

Step 1: Understanding f(x)

The function \( f(x) = \lim_{n \rightarrow -\infty } \left(\cos \sqrt{\frac{x}{n}}\right)^{n} \) requires us to handle limits. As \( n \to -\infty \), it is easier to substitute \( m = -n \) letting \( m \to \infty \). Hence, \( f(x) = \lim_{m \to \infty} \left(\cos \sqrt{\frac{-x}{m}}\right)^{-m} \). For small values of \( t = \sqrt{\frac{-x}{m}} \), \( \cos(t) \approx 1 - \frac{t^2}{2} \). Thus, \( \left( \cos \sqrt{\frac{-x}{m}} \right)^{-m} \approx \left( 1 - \frac{x}{2m} \right)^{-m} \) which is similar to the exponential limit formula \( \left(1 - \frac{z}{m}\right)^{-m} \to e^{-z} \). Applying this, the expression evaluates to \( e^{x/2} \).

Step 2: Understanding g(x)

The function \( g(x) = \lim_{n \rightarrow -\infty} (1 + x + x \sqrt[n]{e})^{n} \) also considers limits where \( n \to -\infty \). Similarly, substituting \( m = -n \), where \( m \to \infty \), \( (1 + x + x \sqrt[m]{e})^{-m} \approx (1 + x)^{-m} \) as \( \sqrt[m]{e} \to 1 \) for large \( m \). Using the exponential limit formula \( (1+t/m)^m \to e^t \), this simplifies to \( e^{-x} \) when \( m \to \infty \).

Step 3: Evaluate limits of logarithms

The limit \( \lim_{x \to 0} \frac{\ln(f(x))}{\ln(g(x))} \) requires calculating \( \ln(e^{x/2}) = \frac{x}{2} \) and \( \ln(e^{-x}) = -x \). Thus, \( \lim_{x \to 0} \frac{\frac{x}{2}}{-x} = \lim_{x \to 0} \frac{x}{2} \cdot \left(-\frac{1}{x}\right) \), which simplifies to \( -\frac{1}{2} \).

Step 4: Conclusion

The evaluated limit is \( \lim_{x \to 0} \frac{\ln(f(x))}{\ln(g(x))} = -\frac{1}{2} \), which matches option (b). Therefore, the answer is \(-\frac{1}{2}\).

Key Concepts

Limits of functions

Limits are a fundamental concept in differential calculus. They help us understand how a function behaves as its input approaches a certain value. When dealing with limits of functions, like in the exercise above, we analyze how functions behave as variables steer towards a point, sometimes infinity, or negative infinity.

In our specific problem, we need to evaluate two separate functions—\(f(x)\) and \(g(x)\)—as \(n\) tends towards negative infinity. By doing this, we aim to determine the behavior of these functions at this extremity. By substituting variables (e.g., transforming \(n\) into \(m = -n\), as seen in the solution), we can better navigate complex expressions and apply limits effectively. This transformation is crucial for making calculations manageable and for veering off into more comprehensible expressions.

Ultimately, identifying the behavior of functions at these critical points allows students to solve otherwise challenging calculus problems and to understand the broader implications in mathematical analysis.

Exponential functions

Exponential functions, characterized by a constant raising to a variable power, play a pivotal role in this problem. Understanding their properties can simplify parts of the limit-taking process. Exponential functions often appear in the limit formulas, such as when transforming an expression like \((1 + \/ x)^{n}\) into an expression involving \(e^t\).

For instance, in our problem, expressing \(f(x)\) and \(g(x)\) in exponential terms makes evaluating the limits possible. It leverages a key property: if \((1 + z/m)^m\) approximates \(e^z\) as \(m\) becomes very large, it clarifies how similar base expressions will behave.

Knowing these properties enables students to refactor seemingly overwhelming expressions into tractable ones, simplifying subsequent mathematical manipulations. Familiarity with exponential growth and decay is not only crucial in calculus but also finds applications across the sciences and engineering.

Logarithmic limits

When working with limits involving logarithms, we often deal with transformations from exponential forms. Logarithms serve to undo exponentials, yielding more approachable expressions. In our solution, it involves using \(\ln(f(x))\) and \(\ln(g(x))\), where these logarithmic rules help simplify to find easier limits.

In the exercise, the logarithmic forms of these functions led to simpler expressions, namely \(\ln(e^{x/2})\) and \(\ln(e^{-x})\), which reflect the power of logarithms in simplifying exponential expressions. Analyzing the ratio of these logs as \(x\) approaches zero ultimately provides the final limit value, demonstrating their utility in comparing function growth or decay rates.

The beauty of logarithms is how they allow students to dissect multiplicative processes into additive ones, a core technique in calculus calculations. As such, logarithmic limits are an extraordinary tool in our mathematical kit.

Inverse trigonometric functions

Inverse trigonometric functions, like \(\tan^{-1}(x)\) used in this problem's context, are crucial for understanding angles based on trigonometric ratios. These functions revert the actions of their corresponding trigonometric functions. In calculus, it's essential to be comfortable with these inverses as they often appear in integration and differentiation problems.

Here, the inverse trigonometric function is part of constructing \(h(x)\). Although it does not directly affect evaluating the given limit \( \lim_{x o 0} \, \frac{\ln (f(x))}{\ln (g(x))} \), its understanding underpins broader concepts where complex functions are constructed based on simpler functional frameworks.

Getting a grasp of these functions means you can transition between angles and ratios effortlessly, which is vital not only in calculus but also in fields like physics, where rotational dynamics might be involved.