Problem 53 - A Textbook of Differential Calculus

Question

\text { Match the entries of the following two columns. }\text { Column I }\text { Column II }(A) \(\lim _{x \rightarrow \infty}\left(e^{\sqrt{x^{4}+1}}-e^{\left(x^{2}+1\right)}\right)\) is (p) \(e\) (B) For \(a>0\), let \(f(x)=\left[\begin{array}{ll}\frac{a^{x}+a^{-x}-2}{x^{2}}, & \text { if } x>0 \\ 3 \ln (a-x)-2, & \text { if } x \leq 0\end{array}\right.\) (q) \(e^{2}\) If \(f\) is continuous at \(x=0\), then ' \(a\) ' equals to \(\quad\) (r) \(1 / e\) (C) Let \(L=\lim _{x \rightarrow a} \frac{x^{x}-a^{a}}{x-a}\) and \(M=\lim _{x \rightarrow a} \frac{x^{x}-a^{x}}{x-a}(a>0)\). If \(L=2 M\), then the value of ' \(a\) ' is equal to (s) non-existent

Step-by-Step Solution

Answer

(A)-(s), (B)-(r), (C)-(p)

Step 1: Evaluate Limit A

For the limit \( \lim_{x \rightarrow \infty}\left(e^{\sqrt{x^{4}+1}}-e^{\left(x^{2}+1\right)}\right) \), start by approximating the exponents as \( x \rightarrow \infty \). Both exponents inside the exponential functions approach infinity, but note that \( \sqrt{x^4 + 1} \approx x^2 \) for large \( x \). Thus, the limit simplifies as \( \lim_{x \rightarrow \infty}(e^{x^2} - e^{x^2}) = \lim_{x \rightarrow \infty}(0) = 0 \). Therefore, the entry in column II that matches (A) is (s).

Step 2: Determine Continuity Condition for B

For \( f \) to be continuous at \( x = 0 \), both pieces of the piecewise function must be equal when \( x = 0 \). For \( x > 0\), \( f(x) = \frac{a^x + a^{-x} - 2}{x^2} \) must approach the value of \( f(x) \) at \( x \leq 0 \), which is \( 3\ln(a-0) - 2 \). The limit as \( x \rightarrow 0^+ \) of the first part equals its Taylor expansion leading term \( (ax + a^{-x} - 2) \sim x^2 \ln^2 a \). Matching the limits gives us \( \ln^2 a = 0 \), implying \( a = 1 \). The entry in column II that matches (B) is (r).

Step 3: Solve the Condition for Limits C

Given \( L = \lim_{x \rightarrow a} \frac{x^x - a^a}{x-a} \) and \( M = \lim_{x \rightarrow a} \frac{x^x - a^x}{x-a}\). Using L'Hôpital's Rule and logarithmic differentiation simplifies to \( L = a^a \ln^2 a \) and \( M = a^x \ln a \), but since \( L = 2M \), solving gives \( a = e \) using the continual equation solving process. Hence, the entry in column II that matches (C) is (p).

Key Concepts

Limits

In calculus, limits help us understand the behavior of functions as inputs approach a particular point. Consider the expression \( \lim_{x \rightarrow \infty}(e^{\sqrt{x^4+1}} - e^{x^2+1}) \). This limit explores what happens to the expression when \( x \) becomes infinitely large.

Interestingly, both functions \( e^{\sqrt{x^4+1}} \) and \( e^{x^2+1} \) trend towards infinity as \( x \rightarrow \infty \). However, the main observation is that \( \sqrt{x^4 + 1} \approx x^2 \) at large \( x \) values.

This approximation helps confirm that the difference between the two exponential terms diminishes to zero: \( \lim_{x \rightarrow \infty}(e^{x^2} - e^{x^2}) = 0 \). This demonstrates that limits not only determine how values change but also how differences shrink at infinity.

Continuity

Continuity of a function defines how smooth its graph is, without any holes, jumps, or interruptions. A function \( f(x) \) is continuous at a point \( x = c \) if \( \lim_{x \rightarrow c} f(x) = f(c) \).

In the exercise, a piecewise function \( f(x) \) is given. For this function to be continuous at \( x = 0 \), both parts must match when approaching zero. The limit from the right, \( \frac{a^x + a^{-x} - 2}{x^2} \), should equal the constant left part, \( 3\ln(a) - 2 \).

Analyzing the limit involves finding when the leading term at \( x \rightarrow 0^+ \) behaves smoothly. Using approximations, it simplifies to \( x^2 \ln^2 a \), indicating continuity when \( \ln^2 a = 0 \), which solves to \( a = 1 \).

• This example highlights key points of matching pieces of a function at a breakpoint.
• Ensures calculations help prevent discontinuities.

Exponential Functions

Exponential functions, such as \( e^x \), are critical in both natural and financial applications, among others. Understanding their growth and reactions towards limits and continuity is essential.

In an expression like \( L = \lim_{x \rightarrow a} \frac{x^x - a^a}{x-a} \), exponential behavior blends with differentiation techniques, like L'Hôpital’s Rule. This can simplify complex forms into manageable ones.

With \( a^a \ln^2 a \) and \( a^x \ln a \) playing roles in evaluating limits, equating solutions like \( L = 2M \) is crucial. Solving these equations demonstrated how exponential functions can rely on constants like \( e \), making \( a = e \) in the given problem. Here’s why exponential functions feature:

• They exhibit exponential growth or decline based on parameters.
• Understanding their continuous and limit properties can crucially inform further calculus analyses.