Question

Let \(f(x)=\frac{e^{\tan x}-e^{x}+\ell n(\sec x+\tan x)-x}{\tan x-x}\), \(x \neq 0\) be a continuous function at \(x=0 .\) The value of \(f(0)\) is equal to (1) \(\frac{1}{2}\) (2) \(\frac{2}{3}\) (3) \(\frac{3}{2}\) (4) 2 (5) \(\frac{4}{3}\)

Short answer

The value of \( f(0) \) is \( \frac{1}{2} \).

Step-by-step solution

Understanding the problem

We need to determine the value that makes the function \( f(x)=\frac{e^{\tan x}-e^{x}+\ln(\sec x+\tan x)-x}{\tan x-x} \) continuous at \( x = 0 \). This implies the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \).

Expand the function near x = 0

We use the series expansions for small angles: \( \tan x \approx x + \frac{x^3}{3} \), \( e^{x} \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} \), and \( \ln(1 + y) \approx y - \frac{y^2}{2} \) for small \( y \).

Applying series expansions

Substitute series expansions near \( x = 0 \) into the function: - \( e^{\tan x} \approx 1 + \tan x + \frac{(\tan x)^2}{2} + \frac{(\tan x)^3}{6} \) - \( e^{x} \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \) - \( \sec x + \tan x \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{3} \)

Substitute and simplify

After substituting and simplifying, we analyze: \( e^{\tan x} - e^x + \ln(\sec x + \tan x) - x \) all divided by \( \tan x - x \). Applying limits, noticing that higher-order terms cancel or become negligibly small:

Calculate limit

Evaluating the limit \( \lim_{{x \to 0}} \frac{1}{2}(x^2 \text{ terms cancel out}) = \frac{1}{2} \).

Key concepts

Continuity of Functions

Continuity is a fundamental concept in calculus. A function is continuous at a specific point if its limit as the variable approaches that point equals the function's value at the point. Mathematically, for a function \( f(x) \), it is continuous at \( x = a \) if \( \text{lim}_{{x \to a}} f(x) = f(a) \).

Regarding our exercise, the function \( f(x) \) needs to be continuous at \( x = 0 \). To find \( f(0) \), we have to ensure that the limit of \( f(x) \) as \( x \) approaches 0 is determined and equals \( f(0) \).

This involves simplifying complex expressions using series expansions near the point and evaluating limits to find the appropriate value.

Series Expansion

Series expansion is the representation of functions as the sum of their terms. Taylor series is often used for this. It helps to approximate functions near a specific point using polynomials.

Here are the series expansions used in the solution:

• \( e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \text{higher-order terms} \)
• \( \tan x \approx x + \frac{x^3}{3} + \text{higher-order terms} \)
• \( \text{ln}(1+y) \approx y - \frac{y^2}{2} + \text{higher-order terms} \)

In the problem provided, we adapt these expansions to expressions like \( e^{\tan(x)} \), \( \text{ln}(\text{sec}(x) + \tan x) \), and \( e^x \) near \( x=0 \). This makes the function easier to handle.

Limits and Calculus

Limits are crucial for understanding the behavior of functions at specific points. When approaching a point, especially where direct substitution results in an indeterminate form like \( \frac{0}{0} \), limits help in determining the actual value.

In our problem, the given fraction

\( f(x) = \frac{e^{\tan x} - e^x + \text{ln}(\text{sec} x + \tan x) - x}{\tan x - x} \) needs to be evaluated as \( x \) approaches 0.

By substituting the series expansions into the function and simplifying, we effectively remove higher-order terms and simplify the limit calculation. Thus, we find:

• The numerator and the denominator retaining only significant terms to evaluate the limit appropriately.
• Finally, using limits, we find that \( f(0) = \frac{1}{2} \), which ensures continuity at \( x=0 \).