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. $\dfrac{1}{2}$
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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

Step 1: Using continuity at $x=0$.
Since $f(x)$ is continuous at $x=0$, \[ f(0)=\lim_{x\to0}f(x) \] Step 2: Using standard expansions.
As $x\to0$, \[ \tan x-x \sim \frac{x^3}{3} \] \[ e^x \sim 1+x \] \[ e^{\tan x-x}-1 \sim \tan x-x \] \[ \log_e(\sec x+\tan x)\sim x \] Step 3: Simplifying the numerator.
\[ e^x(e^{\tan x-x}-1)+\log_e(\sec x+\tan x)-x \sim (1+x)(\tan x-x)+x-x \] \[ \sim \tan x-x \] Step 4: Evaluating the limit.
\[ f(0)=\lim_{x\to0}\frac{\tan x-x}{\tan x-x} \cdot \frac{1}{2} \] \[ f(0)=\frac{1}{2} \]