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

The limit as x approaches-2-of (x^2)/(2x+4) is -infinity

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

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

- \infty

Solution steps

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

Factor out

2 x

from

2 x + 4 : 2 x (1 + \frac{4}{2 x})

= x \to - 2 - lim (\frac{x ^{2}}{2 x ( 1 + \frac{4}{2 x} )})

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

= \frac{1}{2} \cdot x \to - 2 - lim (\frac{x ^{2}}{x ( 1 + \frac{4}{2 x} )})

Simplify

\frac{x ^{2}}{x ( 1 + \frac{4}{2 x} )} : \frac{x ^{2}}{x + 2}

= \frac{1}{2} \cdot x \to - 2 - lim (\frac{x ^{2}}{x + 2})

\frac{x ^{2}}{x + 2} = x^{2} \frac{1}{x + 2}

= \frac{1}{2} \cdot x \to - 2 - lim (x^{2} \frac{1}{x + 2})

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

With the exception of indeterminate form

= \frac{1}{2} \cdot x \to - 2 - lim (x^{2}) \cdot x \to - 2 - lim (\frac{1}{x + 2})

x \to - 2 - lim (x^{2}) = 4

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

= \frac{1}{2} \cdot 4 (- \infty)

Apply Infinity Property:

c \cdot (- \infty) = - \infty

= - \infty

x \to 0 lim (2 sin (3 x))

x \to 0 + lim (x ln (7 x))

x \to 0 lim (- 9)

x \to 2 - lim (\frac{x ^{2} + 5}{x + 1})

x \to 8 + lim (\frac{1}{8 - x})

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