What is the limit as x approaches 9 of ((x^{1/2}-3)(x+1))/(x^2-8x-9) ?

The limit as x approaches 9 of ((x^{1/2}-3)(x+1))/(x^2-8x-9) is 1/6

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limit as x approaches 9 of ((x^{1/2}-3)(x+1))/(x^2-8x-9)

x \to 9 lim \frac{( x ^{\frac{1}{2}} - 3 ) ( x + 1 )}{x ^{2} - 8 x - 9}

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

x \to 9 lim \frac{( x ^{\frac{1}{2}} - 3 ) ( x + 1 )}{x ^{2} - 8 x - 9}

Multiply by the conjugate of

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

= x \to 9 lim \frac{\frac{x - 9}{x ^{\frac{1}{2}} + 3} ( x + 1 )}{x ^{2} - 8 x - 9}

Simplify

\frac{\frac{x - 9}{x ^{\frac{1}{2}} + 3} ( x + 1 )}{x ^{2} - 8 x - 9} : \frac{1}{x ^{\frac{1}{2}} + 3}

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

Plug in the value

x = 9

= \frac{1}{9 ^{\frac{1}{2}} + 3}

Simplify

\frac{1}{9 ^{\frac{1}{2}} + 3} : \frac{1}{6}

= \frac{1}{6}

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What is the limit as x approaches 9 of ((x^{1/2}-3)(x+1))/(x^2-8x-9) ?
The limit as x approaches 9 of ((x^{1/2}-3)(x+1))/(x^2-8x-9) is 1/6