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

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

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

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

3

Solution steps

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

\frac{ln ( e ^{3 x} + x ^{2} )}{x} = ln (e^{3 x} + x^{2}) \frac{1}{x}

= x \to \infty lim (ln (e^{3 x} + x^{2}) \frac{1}{x})

Simplify

ln (e^{3 x} + x^{2}) \frac{1}{x} : \frac{ln ( e ^{3 x} + x ^{2} )}{x}

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{3 e ^{3 x} + 2 x}{e ^{3 x} + x ^{2}}}{1})

= x \to \infty lim (\frac{\frac{3 e ^{3 x} + 2 x}{e ^{3 x} + x ^{2}}}{1})

Simplify

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{9 e ^{3 x} + 2}{e ^{3 x} \cdot 3 + 2 x})

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{27 e ^{3 x}}{9 e ^{3 x} + 2})

= x \to \infty lim (\frac{27 e ^{3 x}}{9 e ^{3 x} + 2})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{81 e ^{3 x}}{27 e ^{3 x}})

= x \to \infty lim (\frac{81 e ^{3 x}}{27 e ^{3 x}})

\frac{81 e ^{3 x}}{27 e ^{3 x}} = 3

= x \to \infty lim (3)

x \to a lim c = c

= 3

x \to - 1 lim (5 x^{2} + 3 x - 4)

x \to 0 lim (\frac{x ^{2} + 6 x}{x})

x \to 0 lim (x^{2} e^{x^{- 2}})

x \to \frac{1}{2} lim (\frac{5}{4})

x \to 1 + lim (1 - x)

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