If $g\left(x\right)=\underset{x\to \infty }{\lim }\frac{x^{m}f\left(1\right)+h\left(x\right)+1}{2x^m+3x+3}$ is continuous at $x=1$ and $g\left(1\right)=\underset{x\to 1}{\lim }{\left\{\ln \left(ex\right)\right\}}^{\frac{2}{\ln x}}$ then find the value of $2g\left(1\right)+2f\left(1\right)-h\left(1\right)$ assume that $f\left(x\right)$ and $h\left(x\right)$ are continuous at $x=1$

Here, $g\left(1\right)=\underset{x\to l}{\lim }{(\ln e+\ln x)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\ln x$}\right.}=\underset{x\to l}{\lim }{\{1+\ln x\}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\ln x$}\right.}=e^{\underset{x\to l}{\lim }\ln x^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\ln x$}\right.}}$

$g\left(1\right)=e^2$

Now, $\underset{x\to 1^{-}}{\lim }g\left(x\right)=\underset{x\to 1^{-}}{\lim }\underset{m\to \infty }{\lim }\left\{\frac{x^{m}f\left(1\right)+h\left(x\right)+1}{2x^m+3x+3}\right\}=\frac{h\left(1\right)+1}{3+3}\left\{\sin cex<1\Rightarrow \underset{m\to \infty }{\lim }{x}^m-0\right\}$

$\therefore \underset{x\to 1^{-}}{\lim }g\left(x\right)=\frac{h\left(1\right)+1}{6}-----\left(2\right)$

and, $\underset{x\to 1^{+}}{\lim }g\left(x\right)=\underset{x\to 1^{+}}{\lim }\underset{m\to \infty }{\lim }\left\{\frac{x^{m}f\left(1\right)+h\left(x\right)+1}{2x^m+3x+3}\right\}=\underset{x\to 1^{+}}{\lim }\underset{m\to \infty }{\lim }\frac{f\left(1\right)+h\left(x\right)/x^m+1/x^m}{2+3/x^{m-1}+3/x^m}=\frac{f\left(1\right)}{2}$

$\therefore \underset{x\to 1^{+}}{\lim }g\left(x\right)=\frac{f\left(1\right)}{2}------\left(3\right)$

As $g\left(x\right)$ is continuous at x = 1, from (1), (2) and (3) 

$e^2=\frac{h\left(1\right)+1}{2}+\frac{f\left(1\right)}{2}\Rightarrow h\left(1\right)=6e^2-1$ and $f\left(1\right)=2e^2$

$\Rightarrow 2g\left(1\right)+2f\left(1\right)-h\left(1\right)=2e^2+4e^2-6e^2+1=1$

$\therefore 2g\left(1\right)+2f\left(1\right)-h\left(1\right)=1$