Question

If the function \[ f(x)=\frac{e^x\left(e^{\tan x - x}-1\right)+\log_e(\sec x+\tan x)-x}{\tan x-x} \] is continuous at $x=0$, then the value of $f(0)$ is equal to

• A. $\dfrac{2}{3}$
• B. $\dfrac{3}{2}$
• C. $2$
• D. $\dfrac{1}{2}$

Hint:

For continuity problems at a point, always replace the function value by the limit and use standard series expansions.

Answer 1

The Correct Option is D

To solve the problem of determining the value of \( f(0) \) for the given function \[ f(x) = \frac{e^x(e^{\tan x - x} - 1) + \log_e(\sec x + \tan x) - x}{\tan x - x} \] when it is continuous at \( x = 0 \), we should first understand the concept of continuity at a point.

A function is continuous at a point if the limit as the variable approaches the point is equal to the function's value at that point. Mathematically, the function \( f(x) \) is continuous at \( x = 0 \) if and only if:

\(\lim_{x \to 0} f(x) = f(0\))

Let's find the limit of \( f(x) \) as \( x \to 0 \).

Step 1: Simplify the Expression

The given function can be split into simpler components:

• \( A(x) = e^x(e^{\tan x - x} - 1) \)
• \( B(x) = \log_e(\sec x + \tan x) - x \)

Hence, \[ f(x) = \frac{A(x) + B(x)}{\tan x - x} \]

Step 2: Evaluate Limits at \( x = 0 \)

We will evaluate the limits of \( A(x) \) and \( B(x) \) separately as \( x \to 0 \).

Limit of \( A(x) \)

For small \( x \), we use the expansion \( e^y \approx 1 + y \) when \( y \) is small:

\(e^{\tan x - x} - 1 \approx \tan x - x\)

Thus, \[ A(x) = e^x(\tan x - x) + o(\tan x - x) \]

Limit of \( B(x) \)

Using the expansion for small angles: \[ \sec x \approx 1 + \frac{x^2}{2}, \quad \tan x \approx x \]

The given expression becomes \[ \log_e(\sec x + \tan x) \approx \log_e(1 + x + \frac{x^2}{2}) \]

Using \( \log_e(1 + y) \approx y \) for small \( y \), \[ \log_e(1 + x + \frac{x^2}{2}) \approx x + \frac{x^2}{2} \]

Thus, \[ B(x) = \left( x + \frac{x^2}{2} \right) - x = \frac{x^2}{2} \]

Step 3: Combine the Limits

The limit of \( f(x) \) as \( x \to 0 \) becomes: \[ \lim_{x \to 0} \frac{e^x(\tan x - x) + \frac{x^2}{2}}{\tan x - x} \]

Using L'Hôpital's rule (since both the numerator and the denominator approach 0 as \( x \to 0 \)), we differentiate the numerator and denominator:

• Numerator's derivative: \( \frac{d}{dx} \left( e^x(\tan x - x) + \frac{x^2}{2} \right) \)
• Denominator's derivative: \( \frac{d}{dx} (\tan x - x) = \sec^2 x - 1 \)

Evaluating these derivatives at \( x = 0 \) gives: \[ \lim_{x \to 0} \frac{e^x(\sec^2 x - 1) + (e^x + e^x (\tan x - x) \frac{d}{dx} (\tan x - x))}{\sec^2 x - 1} = \lim_{x \to 0} \frac{e^x}{e^x} \times \frac{1}{2} = \frac{1}{2} \]

Thus, \( f(0) = \frac{1}{2} \) for the function to be continuous.

Therefore, the value of \( f(0) \) is \(\frac{1}{2}\).