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

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

$$A\left(x\right)=\frac{16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}}{16{(x-4)}^2}letusehospitaltheoremletf\left(x\right)=16\sqrt{x-\sqrt{x}}-3\sqrt{2}x-4\sqrt{2}$$$$andg\left(x\right)=16{(x-4)}^{2}wehave{f}^{{}^{\prime }}\left(x\right)=16\frac{1-\frac{1}{2\sqrt{x}}}{2\sqrt{x-\sqrt{x}}}-3\sqrt{2}$$$$=8\frac{2\sqrt{x}-1}{2\sqrt{x}\sqrt{x-\sqrt{x}}}-3\sqrt{2}=4\times \frac{2\sqrt{x}-1}{\sqrt{x^2-x\sqrt{x}}}-3\sqrt{2}and$$$$f^{\left(2\right)}\left(x\right)=4\frac{\left(\frac{1}{\sqrt{x}}\right)\sqrt{x^2-x\sqrt{x}}-(2\sqrt{x}-1)\times \frac{{(x^2-x\sqrt{x})}^{{}^{\prime }}}{2\sqrt{x^2-x\sqrt{x}}}}{x^2-x\sqrt{x}}$$$${(x^2-x\sqrt{x})}^{{}^{\prime }}=2x-(\sqrt{x}+x\frac{1}{2\sqrt{x}})=2x-(\sqrt{x}+\frac{\sqrt{x}}{2})=2x-\frac{3}{2}\sqrt{x}\Rightarrow$$$$f^{\left(2\right)}\left(x\right)=4\frac{2(x^2-x\sqrt{x})-(2\sqrt{x}-1)(2x-\frac{3\sqrt{x}}{2})}{2\sqrt{x}\sqrt{x^2-x\sqrt{x}}(x^2-x\sqrt{x})}\Rightarrow$$$$f^{\left(2\right)}\left(4\right)=4\times \frac{2(16-8)-\left(3\right)(8-3)}{2.2\sqrt{8}\left(8\right)}=\frac{16-15}{16\sqrt{2}}=\frac{1}{16\sqrt{2}}$$$$g\left(x\right)=16{(x-4)}^2\Rightarrow g^{{}^{\prime }}\left(x\right)=32(x-4)and{g}^{\left(2\right)}\left(x\right)=32\Rightarrow$$$${lim}_{x\to 4}A\left(x\right)=\frac{1}{16\sqrt{2}\times 32}=\frac{1}{16\times 32\sqrt{2}}$$$$$$

Commented by aliesam last updated on 23/Nov/19

$$godblessyousirthankyou$$

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

$$youarewelcome.$$

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