Question

lim x-->infinity (e^3x - e^-3x)/(e^3x + e^-3x) so confused on how to solve

Answer 1

multiply top and bottom by \(e^{-3x}\)

Answer 2

because thats the highest power right?

Answer 3

\[\lim_{x\to\infty}\frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}}\]

\[\lim_{x\to\infty}\frac{e^{3x} - e^{-3x}}{e^{3x} + e^{-3x}}\frac{e^{-3x}}{e^{-3x}}\]

\[\lim_{x\to\infty}\frac{1 - e^{-6x}}{1 + e^{-6x}}=\frac{1+0}{1+0}=1\]

Answer 4

yea i got the same answer too, so you always multiply by the highest power ? what if both top and bottom have the same exponent power?

Answer 5

it all depends on the problem.

Answer 6

ok got it thanks

Answer 7

how about the lim x-> infinity of (sin^2)(x)/(x^2+1) do i also multiply by 1/x^2?