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Question Number 29158 by abdo imad last updated on 04/Feb/18

$$find{lim}_{x\to 1}\frac{\sqrt{3+\sqrt{2x-1}}-2}{\sqrt{2+\sqrt{3x+1}}-\sqrt{x+3}}.$$

Commented by abdo imad last updated on 09/Feb/18

$$inthiscaseitsbettertousehospitaltheoremletput$$$$u\left(x\right)=\sqrt{3+\sqrt{2x-1}}-2andv\left(x\right)=\sqrt{2+\sqrt{3x+1}}-\sqrt{x+3}$$$$u^{{}^{\prime }}\left(x\right)=\frac{\frac{1}{\sqrt{2x-1}}}{2\sqrt{3+\sqrt{2x-1}}}\Rightarrow u^{{}^{\prime }}\left(1\right)=\frac{1}{4}$$$$v^{{}^{\prime }}\left(x\right)=\frac{\frac{3}{2\sqrt{3x+1}}}{2\sqrt{2+\sqrt{3x+1}}}-\frac{1}{2\sqrt{x+3}}\Rightarrow v^{{}^{\prime }}\left(1\right)=\frac{\frac{3}{4}}{4}-\frac{1}{4}$$$$=\frac{3}{16}-\frac{4}{16}=-\frac{1}{16}so$$$${lim}_{x\to 1}\frac{u\left(x\right)}{v\left(x\right)}=\frac{1}{4}\times (-16)=-4.$$$$$$

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