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

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

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

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

Divide by highest denominator power:

\frac{3 + \frac{e ^{- x}}{x}}{2 + \frac{e ^{- 2 x}}{x}}

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

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

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

= \frac{3}{2}

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