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Question Number 141779 by iloveisrael last updated on 23/May/21

$$\underset{x\to 0}{\lim }\frac{x^2}{x+2-2\sqrt{1+x}}=?$$

Answered by Willson last updated on 23/May/21

$$\underset{x\to 0}{\lim }\frac{{\mathrm{x}}^2}{\mathrm{x}+2-2\sqrt{1+\mathrm{x}}}=\underset{x\to 0}{\lim }\frac{2\mathrm{x}}{1-\frac{1}{\sqrt{1+\mathrm{x}}}}$$$$=\underset{x\to 0}{\lim }\frac{2\mathrm{x}\sqrt{1+\mathrm{x}}}{\sqrt{1+\mathrm{x}}-1}$$$$=\underset{x\to 0}{\lim }\frac{2\mathrm{x}\sqrt{1+\mathrm{x}}(\sqrt{1+\mathrm{x}}+1)}{(\sqrt{1+\mathrm{x}}-1)(\sqrt{1+\mathrm{x}}+1)}$$$$=\underset{x\to 0}{\lim }\frac{2\mathrm{x}\sqrt{1+\mathrm{x}}(\sqrt{1+\mathrm{x}}+1)}{\mathrm{x}}$$$$=\underset{x\to 0}{\lim 2}\sqrt{1+\mathrm{x}}(\sqrt{1+\mathrm{x}}+1)$$$$=4$$

Answered by cherokeesay last updated on 23/May/21

$$=\underset{x\to 0}{\lim }\frac{x^2(x+2+2\sqrt{1+x})}{{(x+2)}^2-4(1+x)}=$$$$\underset{x\to 0}{\lim }\frac{x^2(x+2+2\sqrt{1+x})}{x^2+4-4x-4-4x}=2+2=4$$$$$$

Answered by iloveisrael last updated on 23/May/21

Answered by mathmax by abdo last updated on 23/May/21

$$\mathrm{letf}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{\mathrm{x}+2-2\sqrt{1+\mathrm{x}}}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)$$$$={\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{x}}^2(\mathrm{x}+2+2\sqrt{1+\mathrm{x}})}{{(\mathrm{x}+2)}^2-4(1+\mathrm{x})}={\lim }_{\mathrm{x}\to 0}\frac{{\mathrm{x}}^2(\mathrm{x}+2+2\sqrt{1+\mathrm{x}})}{{\mathrm{x}}^2+4\mathrm{x}+4-4-4\mathrm{x}}$$$$={\lim }_{\mathrm{x}\to 0}\mathrm{x}+2+2\sqrt{1+\mathrm{x}}=2+2=4$$

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