What is the limit as x approaches infinity of (x^2ln(x))/(x^{2.01)} ?

The limit as x approaches infinity of (x^2ln(x))/(x^{2.01)} is 0

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limit as x approaches infinity of (x^2ln(x))/(x^{2.01)}

x \to \infty lim (\frac{x ^{2} ln ( x )}{x ^{2.01}})

0

Solution steps

x \to \infty lim (\frac{x ^{2} ln ( x )}{x ^{2.01}})

Simplify

\frac{x ^{2} ln ( x )}{x ^{2.01}} : \frac{ln ( x )}{x ^{0.01}}

= x \to \infty lim (\frac{ln ( x )}{x ^{0.01}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{1}{x}}{\frac{0.01}{x ^{0.99}}})

= x \to \infty lim (\frac{\frac{1}{x}}{\frac{0.01}{x ^{0.99}}})

Simplify

\frac{\frac{1}{x}}{\frac{0.01}{x ^{0.99}}} : \frac{100}{x ^{0.01}}

= x \to \infty lim (\frac{100}{x ^{0.01}})

Apply Infinity Property:

x \to \infty lim (\frac{c}{x ^{a}}) = 0

= 0

x \to 2 lim (\frac{3 x + 2}{x - 1})

x \to 0 lim (x csc (6 x))

x \to 0 lim (sin (x) x)

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

x \to 11 lim (- \frac{10}{x - 11})

What is the limit as x approaches infinity of (x^2ln(x))/(x^{2.01)} ?
The limit as x approaches infinity of (x^2ln(x))/(x^{2.01)} is 0