What is the limit as x approaches infinity of (6e^x+3e^{-x})/(2e^x+4e^{-x)} ?

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limit as x approaches infinity of (6e^x+3e^{-x})/(2e^x+4e^{-x)}

x \to \infty lim (\frac{6 e ^{x} + 3 e ^{- x}}{2 e ^{x} + 4 e ^{- x}})

3

Solution steps

x \to \infty lim (\frac{6 e ^{x} + 3 e ^{- x}}{2 e ^{x} + 4 e ^{- x}})

Expand

\frac{6 e ^{x} + 3 e ^{- x}}{2 e ^{x} + 4 e ^{- x}} : \frac{6 e ^{2 x} + 3}{2 e ^{2 x} + 4}

= x \to \infty lim (\frac{6 e ^{2 x} + 3}{2 e ^{2 x} + 4})

Divide by

e^{2 x} : \frac{6 + \frac{3}{e ^{2 x}}}{2 + \frac{4}{e ^{2 x}}}

= x \to \infty lim (\frac{6 + \frac{3}{e ^{2 x}}}{2 + \frac{4}{e ^{2 x}}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( 6 + \frac{3}{e ^{2 x}} )}{lim _{x \to \infty} ( 2 + \frac{4}{e ^{2 x}} )}

x \to \infty lim (6 + \frac{3}{e ^{2 x}}) = 6

x \to \infty lim (2 + \frac{4}{e ^{2 x}}) = 2

= \frac{6}{2}

Simplify

= 3

What is the limit as x approaches infinity of (6e^x+3e^{-x})/(2e^x+4e^{-x)} ?
The limit as x approaches infinity of (6e^x+3e^{-x})/(2e^x+4e^{-x)} is 3