(A) $\dfrac{2}{3}$

(B) $\dfrac{1}{2}$

(C) $2$

(D) $\dfrac{3}{2}$

Correct Conclusion: Function cannot be continuous at $x=0$

Explanation

For continuity at $x=0$,

$$ \lim_{x\to0} f(x) \quad \text{must be finite} $$

Use standard expansions:

$$ \tan x = x + \frac{x^3}{3} + O(x^5) $$

$$ \sec x = 1 + \frac{x^2}{2} + O(x^4) $$

$$ \log_e(\sec x+\tan x)=x+\frac{x^3}{6}+O(x^5) $$

Denominator:

$$ \tan x - x = \frac{x^3}{3} + O(x^5) $$

Now examine the numerator carefully:

$$ e^{\tan x-x-1}=e^{-1}\left(1+\frac{x^3}{3}+O(x^5)\right) $$

$$ e^x e^{\tan x-x-1} = e^{-1}\left(1+x+\frac{x^2}{2}+\frac{x^3}{2}\right)+O(x^4) $$

Adding remaining terms:

$$ \text{Numerator} = \underbrace{e^{-1}}_{\text{constant}} + e^{-1}x + \frac{e^{-1}}{2}x^2 + \left(\frac{e^{-1}}{2}+\frac{1}{6}\right)x^3 + \cdots $$

Since the denominator is of order $x^3$ but the numerator has a non-zero constant term,

$$ \lim_{x\to0} f(x) = \infty $$

Hence the function cannot be continuous at $x=0$.

Therefore, the given question is internally inconsistent and none of the options is mathematically valid.