What is the limit as x approaches infinity+of sqrt(25x^4+9x)-5x^2 ?

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limit as x approaches infinity+of sqrt(25x^4+9x)-5x^2

x \to \infty + lim (25 x^{4} + 9 x - 5 x^{2})

0

Solution steps

x \to \infty lim (25 x^{4} + 9 x - 5 x^{2})

Multiply by the conjugate of

25 x^{4} + 9 x - 5 x^{2} : \frac{9 x}{25 x ^{4} + 9 x + 5 x ^{2}}

= x \to \infty lim (\frac{9 x}{25 x ^{4} + 9 x + 5 x ^{2}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 9 \cdot x \to \infty lim (\frac{x}{25 x ^{4} + 9 x + 5 x ^{2}})

Divide by highest denominator power:

\frac{\frac{1}{x}}{25 + \frac{9}{x ^{3}} + 5}

= 9 \cdot x \to \infty lim \frac{\frac{1}{x}}{25 + \frac{9}{x ^{3}} + 5}

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

= 9 \cdot \frac{lim _{x \to \infty} ( \frac{1}{x} )}{lim _{x \to \infty} ( 25 + \frac{9}{x ^{3}} + 5 )}

x \to \infty lim (\frac{1}{x}) = 0

x \to \infty lim (25 + \frac{9}{x ^{3}} + 5) = 10

= 9 \cdot \frac{0}{10}

Simplify

9 \cdot \frac{0}{10} : 0

= 0

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What is the limit as x approaches infinity+of sqrt(25x^4+9x)-5x^2 ?
The limit as x approaches infinity+of sqrt(25x^4+9x)-5x^2 is 0