If $x$  is nearly equal to 1 then value of $\frac{m{x}^m-n{x}^n}{m-n}$  is nearly equal to

$x=1+h$

$\frac{m{\left(1+h\right)}^m-n{\left(1+h\right)}^n}{m-n}=\frac{m\left(1+hm\right)-n\left(1+hn\right)}{m-n}$

$=\frac{m+m^{2}h-n-n^{2}h}{m-n}$

$=\frac{\left(m-n\right)+\left(m^2-n^2\right)h}{m-n}$

$=\frac{\left(m-n\right)\left[1+\left(m+n\right)h\right]}{m-n}$

$={\left(1+h\right)}^{m+n}$

$=x^{m+n}$