Question

Let f be a continuous function and g a bounded function such that $f\left(1\right)$ = 297 then find the value of
$\underset{x\to 1}{lim}\underset{n\to \infty }{lim}\frac{f\left(x\right)e^{nx}+g\left(x\right)}{e^{nx}+1}$

Answer 1

$\underset{x\to 1}{lim}\left(\underset{n\to \infty }{lim}\frac{f\left(x\right)e^{nx}+g\left(x\right)}{e^{nx}+1}\right)$

$=\underset{x\to 1}{lim}\left(\underset{n\to \infty }{lim}\left(\frac{f\left(x\right)+g\left(x\right)/e^{nx}}{1+\frac{1}{e^{nx}}}\right)\times \frac{e^{nx}}{e^{nx}}\right)$

$=\underset{x\to 1}{lim}\frac{f\left(x\right)+g\left(x\right)/\infty }{1+1/\infty }=\underset{x\to 1}{lim}f\left(x\right)$

$=f\left(1\right)=297$