Question

$g\left(x\right)={lim}_{m\to \infty }\frac{x^{m}f\left(x\right)+h\left(x\right)+3}{2x^m+4x+1}$ when $x\ne 1$ and $g\left(1\right)=e^3$ such that $f\left(x\right),g\left(x\right)$ and $h\left(x\right)$ are continuous function at $x=1$ and $f\left(1\right)-h\left(1\right)=a(b-g(1\left)\right)$ then $a+b$ is

Answer 1

${lim}_{x\to 1^{+}}g\left(x\right)$ =

${lim}_{m\to \infty }\frac{x^{m}f\left(x\right)+h\left(x\right)+3}{2x^m+4x+1}$

${lim}_{m\to \infty }{lim}_{x\to 1^{+}}\left\{\frac{x^{m}f\left(x\right)+h\left(x\right)+3}{2x^m+4x+1}\right\}$

${lim}_{m\to \infty }{lim}_{m\to 1^{+}}\left\{\frac{f\left(x\right)+\frac{h\left(x\right)+3}{x^m}}{2+\frac{4x+1}{x^m}}\right\}$

$g\left(1\right)=\frac{f\left(1\right)}{2}$

Similarly

$g\left(1\right)={lim}_{x\to 1^{-}}g\left(x\right)=\frac{h\left(1\right)+3}{5}$

$f\left(1\right)=2g\left(1\right)$

$h\left(1\right)=5g\left(1\right)-3$

$f\left(1\right)-h\left(1\right)=-3g\left(1\right)+3$

$3[1-g(1\left)\right]$

Ans.$a=3,b=1,a+b=4$