What is the limit as x approaches 2 of (x^3+4x^2)/(x^2+2x-8) ?

The limit as x approaches 2 of (x^3+4x^2)/(x^2+2x-8) is diverges

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limit as x approaches 2 of (x^3+4x^2)/(x^2+2x-8)

x \to 2 lim (\frac{x ^{3} + 4 x ^{2}}{x ^{2} + 2 x - 8})

diverges

Solution steps

x \to 2 lim (\frac{x ^{3} + 4 x ^{2}}{x ^{2} + 2 x - 8})

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

then the limit does not exist

x \to 2 + lim (\frac{x ^{3} + 4 x ^{2}}{x ^{2} + 2 x - 8}) = \infty

x \to 2 - lim (\frac{x ^{3} + 4 x ^{2}}{x ^{2} + 2 x - 8}) = - \infty

= diverges

h \to 0 lim (\frac{4 - h}{2 + 7 h})

x \to 1 - lim (\frac{x ^{4} - 1}{x - 1})

x \to - 3 lim ((5 x - 1) (2 x + 3))

x \to 2 + lim (e^{\frac{5}{2 - x}})

x \to 0 lim (x sin (\frac{1}{x ^{3}}))

What is the limit as x approaches 2 of (x^3+4x^2)/(x^2+2x-8) ?
The limit as x approaches 2 of (x^3+4x^2)/(x^2+2x-8) is diverges