What is the limit as x approaches 3 of (x+1)/(x^2-9) ?

The limit as x approaches 3 of (x+1)/(x^2-9) is diverges

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limit as x approaches 3 of (x+1)/(x^2-9)

x \to 3 lim (\frac{x + 1}{x ^{2} - 9})

diverges

Solution steps

x \to 3 lim (\frac{x + 1}{x ^{2} - 9})

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

then the limit does not exist

x \to 3 + lim (\frac{x + 1}{x ^{2} - 9}) = \infty

x \to 3 - lim (\frac{x + 1}{x ^{2} - 9}) = - \infty

= diverges

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

\frac{d}{d x} (- \frac{1}{64} sin (\frac{x}{8}))

\frac{d}{d t} (a cos (wt) + b sin (wt))

\int (e^{5 x}) (cos (- 2 x)) d x

\frac{d y}{d x} = y - x, y (0) = \frac{2}{3}

What is the limit as x approaches 3 of (x+1)/(x^2-9) ?
The limit as x approaches 3 of (x+1)/(x^2-9) is diverges