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

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

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

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

3

Solution steps

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{3 e ^{3 x} + 1}{e ^{3 x} + x}}{1})

= x \to \infty lim (\frac{\frac{3 e ^{3 x} + 1}{e ^{3 x} + x}}{1})

Simplify

= x \to \infty lim (\frac{3 e ^{3 x} + 1}{e ^{3 x} + x})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{9 e ^{3 x}}{e ^{3 x} \cdot 3 + 1})

= x \to \infty lim (\frac{9 e ^{3 x}}{e ^{3 x} \cdot 3 + 1})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{27 e ^{3 x}}{9 e ^{3 x}})

= x \to \infty lim (\frac{27 e ^{3 x}}{9 e ^{3 x}})

\frac{27 e ^{3 x}}{9 e ^{3 x}} = 3

= x \to \infty lim (3)

x \to a lim c = c

= 3

t \frac{d y}{d t} + 2 y = t^{4}

\frac{\partial}{\partial y} (x ln (y) - y ln (x))

\int \frac{x ^{4} - x ^{3} + 2 x ^{2} - 2 x - 8}{x ^{2} + 4} d x

x \to 4 lim (\frac{x}{x ^{2} - 3 x})

y^{^{'}} = (x + 1) y^{2}

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