Question

If $x>0$ and $g$ is a bounded function, then $\underset{x\to \infty }{lim}\frac{f\left(y\right)e^{nx}+g\left(y\right)}{e^{nx}+1}$ is where $y$ is independent of $x$

• A. $0$
• B. $g\left(y\right)$
• C. $f\left(y\right)$
• D. $-g\left(y\right)$

Answer 1

The correct option is C $f\left(y\right)$

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

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

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

$=\frac{f\left(y\right)}{1+0}+\frac{g\left(y\right)}{\infty }$

$=f\left(y\right)+0=f\left(y\right)$ ...... $[\frac{a}{\infty }=0]$