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

The limit as x approaches 0 of x/((x+3)^2-9) is 1/6

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

x \to 0 lim (\frac{x}{( x + 3 ) ^{2} - 9})

\frac{1}{6}

Decimal

0.16666 \dots

Solution steps

x \to 0 lim (\frac{x}{( x + 3 ) ^{2} - 9})

Simplify

\frac{x}{( x + 3 ) ^{2} - 9} : \frac{1}{x + 6}

= x \to 0 lim (\frac{1}{x + 6})

Plug in the value

x = 0

= \frac{1}{0 + 6}

Simplify

= \frac{1}{6}

x \to \frac{3 π}{4} lim (2 cos (x))

x \to 101 lim (\frac{x}{x + 4})

x \to 2 lim (x^{4} - 6 x^{2})

x \to 2 + lim (\frac{x + 1}{3 x - 6})

t \to 0 lim (cos (4 t) (cos (8 t)))

What is the limit as x approaches 0 of x/((x+3)^2-9) ?
The limit as x approaches 0 of x/((x+3)^2-9) is 1/6