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

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

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

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

diverges

Solution steps

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

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

then the limit does not exist

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

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

= diverges

h \to 0 lim (\frac{h ∣ h ∣}{h})

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

x \to - 5 lim (π)

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

x \to 3 lim (∣ x ∣ + ∣ - x ∣)

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