If $f(x)=\frac{e^{\tan x} -e^x+\ln (\sec x +\tan x)-x}{\tan x -x}$ is a continuous function at $x=0$, find $f(0)$

Question

Using L’Hospital will provide the answer, the process is long and tedious, so I generally like to avoid it in such questions. But I am not able to find an alternative. Can I get a hint for this?

Answer 1

If the function $f(x)$ is continuous at $x=0$, then $f(0)=L=\lim_{x]ro 0} f(x)$. $$f(x)=\frac{e^{\tan x} -e^x+\ln (\sec x +\tan x)-x}{\tan x -x}$$ $$L=\lim_{x\to 0} \lim_{x\to 0}\frac{e^{\tan x}-e^x}{\tan x-x}+\lim_{x\to 0} \frac{\ln(\sec x+\tan x)-x}{\tan x-x}$$ Using $\lim_{z\to 0}\frac{e^z-1}{z}=1$ $$\implies 1+\lim_{x\to 0}\frac{\ln(\tan(x/2+\pi/4)-x}{\tan x- x}$$ Use $\ln(1+z)=z-z^2/2+z^3/3+...$, then $$\implies L= 1+\lim_{x\to 0}\frac{\ln(1+\tan(x/2)-\ln(1-\tan(x/2))-x}{\tan x- x}$$ Use $\ln(1+z)=z-z^2/2+z^3/3+...$, then $$\implies L=1+\lim_{x \to +...0}\frac{2(\tan(x/2)+(1/3)\tan^3(x/2)+...)-x}{\tan x- x}$$ Next, using $\tan z=z+z^3/3+...$, when $z$ is very small, wr get $$\implies L=1+\lim_{x \to +...0}\frac{2(x/2+(x/2)^3/3+(1/3)[(x/2)+.(x/2)^3/3+..)]^3-x}{\tan x- x}$$ $$\implies L=1+\lim_{x \to +...0}\frac{2]x/2+(x/2)^3/3+(1/3)(x/2)^3+O(x^4)]-x}{\tan x- x}$$ $$\implies L=1+\lim_{x \to +...0}\frac{(x+x^3/6+...)-x}{x^3/3+O(x^5)}$$ Finally, we get $$L=\frac{3}{2}$$