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

The limit as h approaches 0 of ((2+h)^2-8)/h is diverges

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limit as h approaches 0 of ((2+h)^2-8)/h

h \to 0 lim (\frac{( 2 + h ) ^{2} - 8}{h})

diverges

Solution steps

h \to 0 lim (\frac{( 2 + h ) ^{2} - 8}{h})

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

then the limit does not exist

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

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

= diverges

\frac{\partial}{\partial x} (ln (x^{2}))

\frac{\partial}{\partial z} (x z - 5 x^{2} y^{3} z^{4})

\int (\frac{3}{x} - \frac{5}{x ^{6}}) d x

\frac{d y}{d x} = \frac{x ^{3} - x ^{2} + x}{x}

y = tan (x) + \frac{1}{3} tan^{3} (x)

What is the limit as h approaches 0 of ((2+h)^2-8)/h ?
The limit as h approaches 0 of ((2+h)^2-8)/h is diverges