0
$\begingroup$

fine the limits :

$$\lim_{ x \to4 }\frac{x\sqrt{x}+2x-5\sqrt{x}-6}{x\sqrt{x}-8}=?$$

my try :

$$\lim_{ x \to4 }\frac{x\sqrt{x}+2x-5\sqrt{x}-6}{x\sqrt{x}-8}=HOP\\\lim_{ x \to4 }\frac{\dfrac{3x+4\sqrt{x}-5}{2\sqrt{x}} }{\dfrac{3\sqrt{x}}{2} }=5/4$$

BUT :I need to solve without l'Hôpital's rule.

  • 1
    Ahhh!!! Those radicals....Please substitute $\sqrt{x}=t$ to begin with. Then proceed with factoring – 2017-02-15

1 Answers 1

3

Let $\sqrt{x} = u$. Then the limit becomes

$$ \lim_{u \to 2} \frac{u^3 + 2u^2 -5u-6}{u^3 - 8}\\ = \lim_{u\to 2}\frac{u^3 - 2u^2 + 4u^2 - 8u + 3u - 6}{(u-2)(u^2 + 2u + 4)} \\ = \lim_{u\to 2}\frac{(u^2 +4u +3)(u-2)}{(u-2)(u^2+2u+4)} \\ = \frac{5}{4}. $$

  • 1
    You could find the GCD of the numerator and the denominator – 2017-02-15
  • 0
    Equivalent approaches. – 2017-02-15
  • 0
    It's $u^2+2u+4$ in the denominator. – 2017-02-15
  • 0
    Gross error. Corrected. – 2017-02-15