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

For continuity at (x=0),
$$f(0)=\lim_{x\to0}\frac{e^x\left(e^{\tan x-x}-1\right)+\log_e(\sec x+\tan x)-x}{\tan x-x}$$

Use expansions near (x=0):
$$\tan x=x+\frac{x^3}{3}+o(x^3)$$
$$\tan x-x=\frac{x^3}{3}+o(x^3)$$

Also,
$$e^{\tan x-x}-1\sim\tan x-x$$

hence
$$e^x(e^{\tan x-x}-1)\sim\tan x-x$$

$$\log(\sec x+\tan x)$$

has expansion
$$\log(\sec x+\tan x)=x+\frac{x^3}{6}+o(x^3)$$

$$\log(\sec x+\tan x)-x=\frac{x^3}{6}+o(x^3)$$

So numerator becomes
$$\left(\frac{x^3}{3}\right)+\left(\frac{x^3}{6}\right)$$
$$=\frac{x^3}{2}$$

Denominator:
$$\frac{x^3}{3}$$
$$f(0)=\lim_{x\to0}\frac{x^3/2}{x^3/3}$$
$$=\frac{1/2}{1/3}$$
$$=\frac{3}{2}$$