What is the limit as x approaches infinity of (ln(x^2-x+1))/(ln(x^{10)+x+1)} ?

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limit as x approaches infinity of (ln(x^2-x+1))/(ln(x^{10)+x+1)}

x \to \infty lim (\frac{ln ( x ^{2} - x + 1 )}{ln ( x ^{10} + x + 1 )})

\frac{1}{5}

Decimal Notation

0.2

Solution steps

x \to \infty lim (\frac{ln ( x ^{2} - x + 1 )}{ln ( x ^{10} + x + 1 )})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{2 x - 1}{x ^{2} - x + 1}}{\frac{10 x ^{9} + 1}{x ^{10} + x + 1}})

= x \to \infty lim (\frac{\frac{2 x - 1}{x ^{2} - x + 1}}{\frac{10 x ^{9} + 1}{x ^{10} + x + 1}})

Divide fractions:

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

= x \to \infty lim (\frac{( 2 x - 1 ) ( x ^{10} + x + 1 )}{( x ^{2} - x + 1 ) ( 10 x ^{9} + 1 )})

Apply the following algebraic property

: a + b = a (1 + \frac{b}{a})

\frac{( 2 x - 1 ) ( x ^{10} + x + 1 )}{( x ^{2} - x + 1 ) ( 10 x ^{9} + 1 )} = \frac{( x ( 2 - \frac{1}{x} ) ) ( x ^{10} ( 1 + \frac{1}{x ^{9}} + \frac{1}{x ^{10}} ) )}{( x ^{2} ( 1 - \frac{1}{x} + \frac{1}{x ^{2}} ) ) ( x ^{9} ( 10 + \frac{1}{x ^{9}} ) )}

= x \to \infty lim (\frac{x ( 2 - \frac{1}{x} ) x ^{10} ( 1 + \frac{1}{x ^{9}} + \frac{1}{x ^{10}} )}{x ^{2} ( 1 - \frac{1}{x} + \frac{1}{x ^{2}} ) x ^{9} ( 10 + \frac{1}{x ^{9}} )})

Simplify

\frac{( x ( 2 - \frac{1}{x} ) ) ( x ^{10} ( 1 + \frac{1}{x ^{9}} + \frac{1}{x ^{10}} ) )}{( x ^{2} ( 1 - \frac{1}{x} + \frac{1}{x ^{2}} ) ) ( x ^{9} ( 10 + \frac{1}{x ^{9}} ) )} : \frac{( - \frac{1}{x} + 2 ) ( \frac{1}{x ^{9}} + \frac{1}{x ^{10}} + 1 )}{( \frac{1}{x ^{2}} - \frac{1}{x} + 1 ) ( \frac{1}{x ^{9}} + 10 )}

= x \to \infty lim (\frac{( - \frac{1}{x} + 2 ) ( \frac{1}{x ^{9}} + \frac{1}{x ^{10}} + 1 )}{( \frac{1}{x ^{2}} - \frac{1}{x} + 1 ) ( \frac{1}{x ^{9}} + 10 )})

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} ( ( - \frac{1}{x} + 2 ) ( \frac{1}{x ^{9}} + \frac{1}{x ^{10}} + 1 ) )}{lim _{x \to \infty} ( ( \frac{1}{x ^{2}} - \frac{1}{x} + 1 ) ( \frac{1}{x ^{9}} + 10 ) )}

x \to \infty lim ((- \frac{1}{x} + 2) (\frac{1}{x ^{9}} + \frac{1}{x ^{10}} + 1)) = 2

x \to \infty lim ((\frac{1}{x ^{2}} - \frac{1}{x} + 1) (\frac{1}{x ^{9}} + 10)) = 10

= \frac{2}{10}

Cancel the common factor:

2

= \frac{1}{5}

\frac{\partial}{\partial x} (3 sin (2 x))

\int \frac{1}{3 - x} d x

\frac{\partial}{\partial y} (2 y^{3} - 3 y^{2} - 12 y + 6)

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0, y (0) = 4

\int x e^{\frac{- x}{2}} d x

What is the limit as x approaches infinity of (ln(x^2-x+1))/(ln(x^{10)+x+1)} ?
The limit as x approaches infinity of (ln(x^2-x+1))/(ln(x^{10)+x+1)} is 1/5