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Question Number 146505 by mathdanisur last updated on 13/Jul/21

$$\underset{x\to 7}{lim}\frac{\sqrt{x-a}-4}{x-7}=bfinda\cdot b=?$$

Answered by ajfour last updated on 13/Jul/21

$$ijustdunnobutletseewhtits$$$$bout..$$$$iflimithastoexist,i.e.$$$$bfinitethen{d}^r\to 0\Rightarrow n^r\to 0$$$$sox-a=16$$$$butx\to 7\Rightarrow a\to -9$$$$butthen$$$$b=\frac{\frac{\partial n^r}{\partial x}}{\frac{\partial d^r}{\partial x}}=\frac{\frac{1}{2\sqrt{x-a}}}{1}$$$$butgainx\to 7,so$$$$\Rightarrow b=\frac{1}{2\sqrt{7-(-9)}}=\frac{1}{8}$$$$nowifyouask$$$$abforget(\cdot ),uhave-\frac{9}{8}$$

Commented by mathdanisur last updated on 13/Jul/21

$$thanksSer$$

Answered by mathmax by abdo last updated on 13/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{x}-\mathrm{a}}-4}{\mathrm{x}-7}\mathrm{changement}\mathrm{x}-7=\mathrm{t}\mathrm{give}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\sqrt{\mathrm{t}+7-\mathrm{a}}-4}{\mathrm{t}}=\mathrm{g}\left(\mathrm{t}\right)(\mathrm{t}\to 0)\Rightarrow$$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{\sqrt{7-\mathrm{a}}\sqrt{1+\frac{\mathrm{t}}{7-\mathrm{a}}}-4}{\mathrm{t}}\sim \frac{\sqrt{7-\mathrm{a}}\times (1+\frac{\mathrm{t}}{2(7-\mathrm{a})})-4}{\mathrm{t}}$$$$\frac{\sqrt{7-\mathrm{a}}-4}{\mathrm{t}}+\frac{1}{2\sqrt{7-\mathrm{a}}}=\mathrm{b}\Rightarrow \sqrt{7-\mathrm{a}}-4=0\Rightarrow 7-\mathrm{a}=16\Rightarrow \mathrm{a}=-9\Rightarrow$$$$\mathrm{b}=\frac{1}{2\sqrt{7+9}}=\frac{1}{8}\Rightarrow \mathrm{ab}=-\frac{9}{8}$$

Commented by mathdanisur last updated on 13/Jul/21

$$thankyouSer$$

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