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

The limit as x approaches infinity of x/((ln(x))^3+2x) is 1/2

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply L'Hopital's Rule

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

Simplify

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

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

Apply L'Hopital's Rule

= x \to \infty lim (\frac{1}{\frac{6 l n ( x )}{x} + 2})

Simplify

\frac{1}{\frac{6 l n ( x )}{x} + 2} : \frac{x}{6 ln ( x ) + 2 x}

= x \to \infty lim (\frac{x}{6 ln ( x ) + 2 x})

Apply L'Hopital's Rule

= x \to \infty lim (\frac{1}{\frac{6}{x} + 2})

Simplify

\frac{1}{\frac{6}{x} + 2} : \frac{x}{6 + 2 x}

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

Divide by highest denominator power:

\frac{1}{\frac{6}{x} + 2}

= x \to \infty lim (\frac{1}{\frac{6}{x} + 2})

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} ( 1 )}{lim _{x \to \infty} ( \frac{6}{x} + 2 )}

x \to \infty lim (1) = 1

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

= \frac{1}{2}

What is the limit as x approaches infinity of x/((ln(x))^3+2x) ?
The limit as x approaches infinity of x/((ln(x))^3+2x) is 1/2