What is the limit as x approaches 0 of 1/(x^2(x+8)) ?

The limit as x approaches 0 of 1/(x^2(x+8)) is infinity

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limit as x approaches 0 of 1/(x^2(x+8))

x \to 0 lim (\frac{1}{x ^{2} ( x + 8 )})

\infty

Solution steps

x \to 0 lim (\frac{1}{x ^{2} ( x + 8 )})

x \to a - lim f (x) = x \to a + lim f (x) = L

then

x \to a lim f (x) = L

x \to 0 - lim (\frac{1}{x ^{2} ( x + 8 )}) = \infty

x \to 0 + lim (\frac{1}{x ^{2} ( x + 8 )}) = \infty

= \infty

y^{^{''}} + 4 y = sin (2 x)

\int_{0}^{8} 3 x d x

\frac{d y}{d x} + 4 y^{2} = 0

t an g e n t f (x) = \frac{6}{x}, at x = 5

\int_{0}^{4} t e^{- t} d t

What is the limit as x approaches 0 of 1/(x^2(x+8)) ?
The limit as x approaches 0 of 1/(x^2(x+8)) is infinity