What is the limit as x approaches 2 of (x^2+x-6)/(x^3-9x) ?

The limit as x approaches 2 of (x^2+x-6)/(x^3-9x) is 0

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limit as x approaches 2 of (x^2+x-6)/(x^3-9x)

x \to 2 lim (\frac{x ^{2} + x - 6}{x ^{3} - 9 x})

0

Solution steps

x \to 2 lim (\frac{x ^{2} + x - 6}{x ^{3} - 9 x})

Plug in the value

x = 2

= \frac{2 ^{2} + 2 - 6}{2 ^{3} - 9 \cdot 2}

Simplify

\frac{2 ^{2} + 2 - 6}{2 ^{3} - 9 \cdot 2} : 0

= 0

x \to 0 lim (1 - cos (6 x))

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

x \to \infty lim (5 x - x^{2})

n \to \infty lim (\frac{ln ( n + 2 )}{ln ( n + 1 )})

x \to 1 lim (sin (- π x))

What is the limit as x approaches 2 of (x^2+x-6)/(x^3-9x) ?
The limit as x approaches 2 of (x^2+x-6)/(x^3-9x) is 0