lim-x-2-x-2-2x-4-x-2-2-x-5-32-no-hospital-

Question Number 132483 by abdullahquwatan last updated on 14/Feb/21

\lim_{x \to - 2} \frac{\sqrt{x^{2} + 2 x + 4} - x^{2} + 2}{x^{5} + 32} no hospital

Commented by EDWIN88 last updated on 14/Feb/21

what hospital ?

Commented by abdullahquwatan last updated on 14/Feb/21

l^{'} hopital theorem

Answered by EDWIN88 last updated on 14/Feb/21

let x + 2 = t \land t \to 0

\lim_{t \to 0} \frac{\sqrt{{(t - 2)}^{2} + 2 (t - 2) + 4} + 2 - {(t - 2)}^{2}}{{(t - 2)}^{5} + 32}

\lim_{t \to 0} \frac{\sqrt{t^{2} - 2 t + 4} - (t^{2} - 4 t + 2)}{{(t - 2)}^{5} + 32} =

\lim_{t \to 0} \frac{(t^{2} - 2 t + 4) - {(t^{2} - 4 t + 2)}^{2}}{{(t - 2)}^{5} + 32} \times \lim_{t \to 0} \frac{1}{\sqrt{t^{2} - 2 t + 4} + t^{2} - 4 t + 2}

= \frac{1}{4} \times \lim_{t \to 0} \frac{t^{2} - 2 t + 4 - (t^{4} + {4 t}^{2} + 4 - {8 t}^{3} + {4 t}^{2} - 16 t)}{t^{5} - {10 t}^{4} + {20 t}^{3} - {80 t}^{2} + 80 t}

= \frac{1}{4} \times \lim_{t \to 0} \frac{- t^{4} + {8 t}^{3} - {7 t}^{2} + 14 t}{t^{5} - {10 t}^{4} + {20 t}^{3} - {80 t}^{2} + 80 t}

= \frac{1}{4} \times \frac{14}{80} = \frac{7}{160}

Commented by abdullahquwatan last updated on 14/Feb/21

t h a n k y o u s o m u c h

Answered by bemath last updated on 14/Feb/21

If by L^{'} H \ddot{o p i t a l} rule

\lim_{x \to - 2} \frac{\frac{x + 1}{\sqrt{x^{2} + 2 x + 4}} - 2 x}{{5 x}^{4}} =

\frac{\frac{- 1}{\sqrt{4}} + 4}{80} = \frac{- \frac{1}{2} + 4}{80} = \frac{- 1 + 8}{160} = \frac{7}{160}

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