lim-x-4-2x-2-16-4-4x-2-16x-1-2x-2-16-4-

Question Number 131388 by Eric002 last updated on 04/Feb/21

\underset{x \to 4}{l i m} \frac{\sqrt{2 x^{2} - 16} + 4}{4 x^{2} - 16 x} - \frac{1}{\sqrt{2 x^{2} - 16} - 4}

Answered by bemath last updated on 04/Feb/21

\lim_{x \to 4} (\frac{\sqrt{2 x^{2} - 16} + 4}{4 x^{2} - 16 x} - \frac{1}{\sqrt{2 x^{2} - 16} - 4}) = ?

(∙) \frac{1}{\sqrt{2 x^{2} - 16} - 4} = \frac{\sqrt{2 x^{2} - 16} + 4}{2 x^{2} - 32}

(∙ ∙) \lim_{x \to 4} \frac{\sqrt{2 x^{2} - 16} + 4}{4 x^{2} - 16 x} - \frac{\sqrt{2 x^{2} - 16} + 4}{2 x^{2} - 32}

\lim_{x \to 4} (\sqrt{2 x^{2} - 16} + 4) \times \lim_{x \to 4} (\frac{1}{4 x^{2} - 16 x} - \frac{1}{2 x^{2} - 32})

= 8 \times \lim_{x \to 4} (\frac{1}{4 x (x - 4)} - \frac{1}{2 (x - 4) (x + 4)})

= 8 \times \lim_{x \to 4} (\frac{x + 4}{4 x (x - 4) (x + 4)} - \frac{2 x}{4 x (x - 4) (x + 4)})

= 8 \times \lim_{x \to 4} (\frac{- (x - 4)}{4 x (x - 4) (x + 4)})

= 8 \times \lim_{x \to 4} (\frac{- 1}{4 x (x + 4)}) = - \frac{8}{16 \times 8}

= - \frac{1}{16}

Commented by bramlexs22 last updated on 04/Feb/21

Wow es ist so cool Sir

Commented by Eric002 last updated on 04/Feb/21

w e l l d o n e s i r

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