Question

Mathematics

Prove that:
Lim {ae^x+ be^(-x)}/{ce^x + de^(-x)}=a/c,when x tends to infinity

Answer 1

lim               ae^x+be^-x
x tends to ∞  ce^x+de^-x
taking common e^x from nominator as well denominator
lim                e^x(a+be^-2x)
x tends to∞    e^x(c+de^-2x)
Apply limit
a+be^-∞          ⇒a/c    ( e^∞=0 )
c+de^-∞

Answer 2

$\lim_{x \to \infty} \frac{a\ e^x+ b\ e^{-x}}{c\ e^x + d\ e^{-x}}\\\\= \lim_{x \to \infty} \frac{e^x(a+ b\ e^{-2x})}{e^x(c + d\ e^{-2x})}\\\\= \lim_{x \to \infty} \frac{a+ b\ e^{-2x}}{c + d\ e^{-2x}}\\\\=\lim_{x \to \infty} \frac{a+ b\ 0}{c + d\ 0},=\frac{a}{c}\\\\ as\ lim_{x \to \infty}\ e^{-2x}}\\=as\ lim_{x \to \infty}\ \frac{1}{e^{2x}}}\\=\frac{1}{\infty}=0$