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

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

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

\frac{3}{2}

Decimal

1.5

Solution steps

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

Apply L'Hopital's Rule

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

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

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

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 9 e ^{3 x} + 5 e ^{5 x} )}{2 ( 4 e ^{2 x} + 5 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 27 e ^{3 x} + 25 e ^{5 x} )}{2 ( 8 e ^{2 x} + 25 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 81 e ^{3 x} + 125 e ^{5 x} )}{2 ( 16 e ^{2 x} + 125 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 243 e ^{3 x} + 625 e ^{5 x} )}{2 ( 32 e ^{2 x} + 625 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 729 e ^{3 x} + 3125 e ^{5 x} )}{2 ( 64 e ^{2 x} + 3125 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 2187 e ^{3 x} + 15625 e ^{5 x} )}{2 ( 128 e ^{2 x} + 15625 e ^{5 x} )})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{3 ( 6561 e ^{3 x} + 78125 e ^{5 x} )}{2 ( 256 e ^{2 x} + 78125 e ^{5 x} )})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{3}{2} \cdot x \to \infty lim (\frac{6561 e ^{3 x} + 78125 e ^{5 x}}{256 e ^{2 x} + 78125 e ^{5 x}})

Divide by

e^{5 x} : \frac{\frac{6561}{e ^{2 x}} + 78125}{\frac{256}{e ^{3 x}} + 78125}

= \frac{3}{2} \cdot x \to \infty lim (\frac{\frac{6561}{e ^{2 x}} + 78125}{\frac{256}{e ^{3 x}} + 78125})

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{3}{2} \cdot \frac{lim _{x \to \infty} ( \frac{6561}{e ^{2 x}} + 78125 )}{lim _{x \to \infty} ( \frac{256}{e ^{3 x}} + 78125 )}

x \to \infty lim (\frac{6561}{e ^{2 x}} + 78125) = 78125

x \to \infty lim (\frac{256}{e ^{3 x}} + 78125) = 78125

= \frac{3}{2} \cdot \frac{78125}{78125}

Simplify

= \frac{3}{2}

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