What is the limit as x approaches infinity of (3x^2+2x-8)/(sqrt(4x^4-5x+4)) ?

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limit as x approaches infinity of (3x^2+2x-8)/(sqrt(4x^4-5x+4))

x \to \infty lim (\frac{3 x ^{2} + 2 x - 8}{4 x ^{4} - 5 x + 4})

\frac{3}{2}

Decimal

1.5

Solution steps

x \to \infty lim (\frac{3 x ^{2} + 2 x - 8}{4 x ^{4} - 5 x + 4})

Divide by highest denominator power:

\frac{3 + \frac{2}{x} - \frac{8}{x ^{2}}}{4 - \frac{5}{x ^{3}} + \frac{4}{x ^{4}}}

= x \to \infty lim \frac{3 + \frac{2}{x} - \frac{8}{x ^{2}}}{4 - \frac{5}{x ^{3}} + \frac{4}{x ^{4}}}

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{2}{x} - \frac{8}{x ^{2}} )}{lim _{x \to \infty} ( 4 - \frac{5}{x ^{3}} + \frac{4}{x ^{4}} )}

x \to \infty lim (3 + \frac{2}{x} - \frac{8}{x ^{2}}) = 3

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

= \frac{3}{2}

x \to 0 lim (\frac{1}{x + 1})

\int (\frac{1}{3}) x^{3} - 5 x d x

\int (6 x^{2} - 4 x + 5) d x

\int \frac{1}{x ^{3} + 9 x} d x

t an g e n t 5 sec (x), at x = \frac{π}{3}

What is the limit as x approaches infinity of (3x^2+2x-8)/(sqrt(4x^4-5x+4)) ?
The limit as x approaches infinity of (3x^2+2x-8)/(sqrt(4x^4-5x+4)) is 3/2