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

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

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

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Divide by highest denominator power:

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

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

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

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

x \to \infty lim (4 + \frac{2}{x}) = 4

= \frac{1}{4}

x \to \infty lim (- \frac{x}{0})

n \to \infty lim (e^{- n + 2})

h \to 0 lim (csc (h) - cot (h))

x \to - 3 lim ((4 x - 2 \cdot 3)^{2})

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