Question

If $f\left(x\right)={\left(\frac{x}{2+x}\right)}^{2x},$ then

• A. ${lim}_{x\to \infty }f\left(x\right)=e^{-6}$
  
• B. ${lim}_{x\to \infty }f\left(x\right)=2$
• C. ${lim}_{x\to \infty }f\left(x\right)=e^{-4}$
• D.
  ${lim}_{x\to \infty }f\left(x\right)=\frac{1}{9}$

Answer 1

Answer: (C), (D)

The correct options are
C ${lim}_{x\to \infty }f\left(x\right)=e^{-4}$
D

${lim}_{x\to \infty }f\left(x\right)=\frac{1}{9}$
${\left(\frac{x}{2+x}\right)}^{2x}={(1+\frac{-2}{2+x})}^{2x}$

$={(1+\frac{-2}{2+x})}^{\frac{2+x}{-2}(\frac{-2}{2+x}\times 2x)}$

Also ${lim}_{x\to \infty }{(1+\frac{-2}{2+x})}^{\frac{2+x}{-2}}={lim}_{t\to 0}(1+t{)}^{\frac{1}{t}}=e$ and ${lim}_{x\to \infty }\frac{-2}{2+x}.2x=-4$

Thus, ${lim}_{x\to \infty }f\left(x\right)=e^{-4}$

Also,

${lim}_{x\to 1}f\left(x\right)={\left(\frac{1}{3}\right)}^2=\frac{1}{9}$