Question-74383

Question Number 74383 by aliesam last updated on 23/Nov/19

Commented by mathmax by abdo last updated on 23/Nov/19

A (x) = \frac{16 \sqrt{x - \sqrt{x}} - 3 \sqrt{2} x - 4 \sqrt{2}}{16 {(x - 4)}^{2}} l e t u s e h o s p i t a l t h e o r e m l e t f (x) = 16 \sqrt{x - \sqrt{x}} - 3 \sqrt{2} x - 4 \sqrt{2}

a n d g (x) = 16 {(x - 4)}^{2} w e h a v e f^{^{'}} (x) = 16 \frac{1 - \frac{1}{2 \sqrt{x}}}{2 \sqrt{x - \sqrt{x}}} - 3 \sqrt{2}

= 8 \frac{2 \sqrt{x} - 1}{2 \sqrt{x} \sqrt{x - \sqrt{x}}} - 3 \sqrt{2} = 4 \times \frac{2 \sqrt{x} - 1}{\sqrt{x^{2} - x \sqrt{x}}} - 3 \sqrt{2} a n d

f^{(2)} (x) = 4 \frac{(\frac{1}{\sqrt{x}}) \sqrt{x^{2} - x \sqrt{x}} - (2 \sqrt{x} - 1) \times \frac{{(x^{2} - x \sqrt{x})}^{^{'}}}{2 \sqrt{x^{2} - x \sqrt{x}}}}{x^{2} - x \sqrt{x}}

{(x^{2} - x \sqrt{x})}^{^{'}} = 2 x - (\sqrt{x} + x \frac{1}{2 \sqrt{x}}) = 2 x - (\sqrt{x} + \frac{\sqrt{x}}{2}) = 2 x - \frac{3}{2} \sqrt{x} \Rightarrow

f^{(2)} (x) = 4 \frac{2 (x^{2} - x \sqrt{x}) - (2 \sqrt{x} - 1) (2 x - \frac{3 \sqrt{x}}{2})}{2 \sqrt{x} \sqrt{x^{2} - x \sqrt{x}} (x^{2} - x \sqrt{x})} \Rightarrow

f^{(2)} (4) = 4 \times \frac{2 (16 - 8) - (3) (8 - 3)}{2.2 \sqrt{8} (8)} = \frac{16 - 15}{16 \sqrt{2}} = \frac{1}{16 \sqrt{2}}

g (x) = 16 {(x - 4)}^{2} \Rightarrow g^{^{'}} (x) = 32 (x - 4) a n d g^{(2)} (x) = 32 \Rightarrow

{l i m}_{x \to 4} A (x) = \frac{1}{16 \sqrt{2} \times 32} = \frac{1}{16 \times 32 \sqrt{2}}

Commented by aliesam last updated on 23/Nov/19

g o d b l e s s y o u s i r t h a n k y o u

Commented by mathmax by abdo last updated on 23/Nov/19

y o u a r e w e l c o m e .