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

解答

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

\frac{17}{18}

十进制

0.94444 \dots

求解步骤

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

乘以

72 - x + x

的共轭根式

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

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

化简

\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})

代入值

x = - 9

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

化简

- \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 )})

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

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