What is the limit as x approaches 0 of ((2x^2+16))/x ?

The limit as x approaches 0 of ((2x^2+16))/x is diverges

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limit as x approaches 0 of ((2x^2+16))/x

x \to 0 lim (\frac{( 2 x ^{2} + 16 )}{x})

diverges

Solution steps

x \to 0 lim (\frac{2 x ^{2} + 16}{x})

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

then the limit does not exist

x \to 0 + lim (\frac{2 x ^{2} + 16}{x}) = \infty

x \to 0 - lim (\frac{2 x ^{2} + 16}{x}) = - \infty

= diverges

x^{2} e^{x^{2}}

n = 1 \sum \infty n \cdot (\frac{3}{4})^{n}

\int \frac{9 - x ^{2}}{x ^{3}} d x

\int \frac{1}{x ^{2} 16 - x ^{2}} d x

What is the limit as x approaches 0 of ((2x^2+16))/x ?
The limit as x approaches 0 of ((2x^2+16))/x is diverges