What is the limit as x approaches infinity of (3x^2-x-2)/(5x^2-4x+1) ?

The limit as x approaches infinity of (3x^2-x-2)/(5x^2-4x+1) is 3/5

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limit as x approaches infinity of (3x^2-x-2)/(5x^2-4x+1)

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

\frac{3}{5}

Decimal Notation

0.6

Solution steps

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

Divide by highest denominator power:

\frac{3 - \frac{1}{x} - \frac{2}{x ^{2}}}{5 - \frac{4}{x} + \frac{1}{x ^{2}}}

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

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

x \to \infty lim (5 - \frac{4}{x} + \frac{1}{x ^{2}}) = 5

= \frac{3}{5}

f (x) = \frac{x ^{2}}{3 + 8 x}

(\frac{x ^{2}}{2})^{^{'}}

\int_{0}^{4} x 17 - x^{2} d x

\int_{0}^{1} \frac{x ^{4} + 6}{x} d x

tan^{4} (2 x - 1)^{3}

What is the limit as x approaches infinity of (3x^2-x-2)/(5x^2-4x+1) ?
The limit as x approaches infinity of (3x^2-x-2)/(5x^2-4x+1) is 3/5