Question

$\underset{y\to \infty }{lim}{\left(\frac{y-3}{y+2}\right)}^y$ is equal to ?

Answer 1

$\underset{y\to \infty }{lim}{\left(\frac{y-3}{y+2}\right)}^y=\underset{y\to \infty }{lim}\left(e^{ylog\frac{y-3}{y+2}}\right)[\because a^x=e^{log{a}^x}=e^{x.loga}][bychainrule,g\left(y\right)=ylog\left(\frac{y-3}{y+2}\right),f\left(u\right)=e^u]\therefore \underset{y\to \infty }{lim}\left(ylog\left(\frac{y-3}{y+2}\right)\right)=-5and,\underset{u\to -5}{lim}\left(e^u\right)=\frac{1}{e^5}\therefore \underset{y\to \infty }{lim}{\left(\frac{y-3}{y+2}\right)}^y=\frac{1}{e^5}$