What is the limit as x approaches 3 of (x-5)/(x^2-9) ?

The limit as x approaches 3 of (x-5)/(x^2-9) is diverges

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limit as x approaches 3 of (x-5)/(x^2-9)

x \to 3 lim (\frac{x - 5}{x ^{2} - 9})

diverges

Solution steps

x \to 3 lim (\frac{x - 5}{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 - 5}{x ^{2} - 9}) = - \infty

x \to 3 - lim (\frac{x - 5}{x ^{2} - 9}) = \infty

= diverges

x \to 0 + lim (\frac{1}{x} sin (\frac{x}{2}))

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

x \to 0 lim (\frac{- sin ( 5 x )}{x})

x \to \infty lim (\frac{3}{4 x})

x \to 0 + lim (\frac{x ^{2}}{e ^{x} - 1})

What is the limit as x approaches 3 of (x-5)/(x^2-9) ?
The limit as x approaches 3 of (x-5)/(x^2-9) is diverges