What is the limit as x approaches-9 of (sqrt(72-x)+x)/(9+x) ?

The limit as x approaches-9 of (sqrt(72-x)+x)/(9+x) is 17/18

limit as x approaches-9 of (sqrt(72-x)+x)/(9+x)

x \to - 9 lim (\frac{72 - x + x}{9 + x})

\frac{17}{18}

Decimal

0.94444 \dots

Solution steps

x \to - 9 lim (\frac{72 - x + x}{9 + x})

Multiply by the conjugate of

72 - x + x : \frac{72 - x - x ^{2}}{72 - x - x}

= x \to - 9 lim \frac{\frac{72 - x - x ^{2}}{72 - x - x}}{9 + x}

Simplify

\frac{\frac{72 - x - x ^{2}}{72 - x - x}}{9 + x} : - \frac{x - 8}{- x + - x + 72}

= x \to - 9 lim (- \frac{x - 8}{- x + - x + 72})

Plug in the value

x = - 9

= - \frac{- 9 - 8}{- ( - 9 ) + - ( - 9 ) + 72}

Simplify

- \frac{- 9 - 8}{- ( - 9 ) + - ( - 9 ) + 72} : \frac{17}{18}

= \frac{17}{18}

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

x \to 0 lim (\frac{1 - cos ( 4 x )}{1 - cos ( 6 x )})

x \to 2 lim (- x + 1)

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

x \to 5 - lim (\frac{( 1 )}{x - 5})

What is the limit as x approaches-9 of (sqrt(72-x)+x)/(9+x) ?
The limit as x approaches-9 of (sqrt(72-x)+x)/(9+x) is 17/18