What is the limit as x approaches infinity of (sqrt(225x^2+7x-23))/(15x) ?

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limit as x approaches infinity of (sqrt(225x^2+7x-23))/(15x)

x \to \infty lim (\frac{225 x ^{2} + 7 x - 23}{15 x})

1

Solution steps

x \to \infty lim (\frac{225 x ^{2} + 7 x - 23}{15 x})

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

= \frac{1}{15} \cdot x \to \infty lim (\frac{225 x ^{2} + 7 x - 23}{x})

Divide by highest denominator power:

225 + \frac{7}{x} - \frac{23}{x ^{2}}

= \frac{1}{15} \cdot x \to \infty lim (225 + \frac{7}{x} - \frac{23}{x ^{2}})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= \frac{1}{15} x \to \infty lim (225 + \frac{7}{x} - \frac{23}{x ^{2}})

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

= \frac{1}{15} x \to \infty lim (225) + x \to \infty lim (\frac{7}{x}) - x \to \infty lim (\frac{23}{x ^{2}})

x \to \infty lim (225) = 225

x \to \infty lim (\frac{7}{x}) = 0

x \to \infty lim (\frac{23}{x ^{2}}) = 0

= \frac{1}{15} 225 + 0 - 0

Simplify

\frac{1}{15} 225 + 0 - 0 : 1

= 1

\int cos (x) x d x

f (x) = (9 - x^{3})^{- 2}

10 y^{^{''}} + 2 y^{^{'}} + 5 y = 0

\frac{d y}{d x} = \frac{4 y}{x}

ln (x + 4)

What is the limit as x approaches infinity of (sqrt(225x^2+7x-23))/(15x) ?
The limit as x approaches infinity of (sqrt(225x^2+7x-23))/(15x) is 1