What is the limit as x approaches 3-of (x+2)/(x^2-9) ?

The limit as x approaches 3-of (x+2)/(x^2-9) is -infinity

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limit as x approaches 3-of (x+2)/(x^2-9)

x \to 3 - lim (\frac{x + 2}{x ^{2} - 9})

- \infty

Solution steps

x \to 3 - lim (\frac{x + 2}{x ^{2} - 9})

Multiply by the conjugate of

x + 2 : \frac{x ^{2} - 4}{x - 2}

= x \to 3 - lim (\frac{\frac{x ^{2} - 4}{x - 2}}{x ^{2} - 9})

Simplify

\frac{\frac{x ^{2} - 4}{x - 2}}{x ^{2} - 9} : \frac{x + 2}{( x + 3 ) ( x - 3 )}

= x \to 3 - lim (\frac{x + 2}{( x + 3 ) ( x - 3 )})

\frac{x + 2}{( x + 3 ) ( x - 3 )} = (x + 2) \frac{1}{( x + 3 ) ( x - 3 )}

= x \to 3 - lim ((x + 2) \frac{1}{( x + 3 ) ( x - 3 )})

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

= x \to 3 - lim (x + 2) \cdot x \to 3 - lim (\frac{1}{( x + 3 ) ( x - 3 )})

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

x \to 3 - lim (\frac{1}{( x + 3 ) ( x - 3 )}) = - \infty

= 5 (- \infty)

Apply Infinity Property:

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

= - \infty

x + 8

\frac{\partial}{\partial x} (e^{\frac{x}{y}})

\frac{\partial}{\partial r} (r sin (t) sin (x))

\int_{0}^{1} \frac{17}{4 y - 1} d y

\frac{6}{s} + 8 a \cdot \frac{s}{s ^{2} + 16}

What is the limit as x approaches 3-of (x+2)/(x^2-9) ?
The limit as x approaches 3-of (x+2)/(x^2-9) is -infinity