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Question Number 143064 by EDWIN88 last updated on 09/Jun/21

$$\underset{x\to 0}{\lim }\frac{\sqrt[3]{1+{\mathrm{x}}^2}-\sqrt[4]{1-2\mathrm{x}}}{\mathrm{x}+{\mathrm{x}}^2}=?$$$$\underset{x\to 1}{\lim }\frac{\sqrt[3]{7+{\mathrm{x}}^2}-\sqrt{3+{\mathrm{x}}^2}}{\mathrm{x}-1}=?$$

Answered by bramlexs22 last updated on 09/Jun/21

Answered by Mathspace last updated on 09/Jun/21

$$f\left(x\right)=\frac{{(1+x^2)}^3-{(1-2x)}^{\frac{1}{4}}}{x^2+x}\Rightarrow$$$$f\left(x\right)\sim \frac{1+3x^2-(1-\frac{x}{2})}{x^2+x}\Rightarrow$$$$f\left(x\right)\sim \frac{3x^2+\frac{x}{2}}{x^2+x}=\frac{3x+\frac{1}{2}}{x+1}\Rightarrow$$$${lim}_{x\to 0}f\left(x\right)=\frac{1}{2}$$

Answered by Mathspace last updated on 09/Jun/21

$$g\left(x\right)=\frac{{(7+x^2)}^{\frac{1}{3}}-{(3+x^2)}^{\frac{1}{2}}}{x-1}$$$$wedothechangementx-1=t(sot\to 0)$$$$g\left(x\right)=g(t+1)=\frac{{(7+{(t+1)}^2)}^{\frac{1}{3}}-{(3+{(t+1)}^2)}^{\frac{1}{2}}}{t}$$$$=\frac{{(7+t^2+2t+1)}^{\frac{1}{3}}-{(3+t^2+2t+1)}^{\frac{1}{2}}}{t}$$$$=\frac{{(8+t^2+2t)}^{\frac{1}{3}}-{(4+t^2+2t)}^{\frac{1}{2}}}{t}$$$$g(t+1)=\frac{2{(1+\frac{t^2+2t}{8})}^{\frac{1}{3}}-2{(1+\frac{t^2+2t}{4})}^{\frac{1}{2}}}{t}$$$$\sim \frac{2(1+\frac{1}{24}(t^2+2t))-2(1+\frac{1}{8}(t^2+2t))}{t}$$$$=\frac{\frac{1}{12}(t^2+2t)-\frac{1}{4}(t^2+2t)}{t}$$$$=-\frac{1}{6}\times \frac{t^2+2t}{t}=-\frac{1}{6}(t+2)\to -\frac{1}{3}$$$$\Rightarrow {lim}_{t\to 1}g\left(x\right)=-\frac{1}{3}$$

Answered by Olaf_Thorendsen last updated on 09/Jun/21

$$f\left(x\right)=\frac{\sqrt[3]{1+x^2}-\sqrt[4]{1-2x}}{x+x^2}$$$$f\left(x\right)\underset{0}{\sim }\frac{(1+\frac{1}{3}{x}^2)-(1+\frac{1}{4}(-2x))}{x+x^2}$$$$f\left(x\right)\underset{0}{\sim }\frac{\frac{1}{3}{x}^2+\frac{1}{2}x}{x+x^2}\to \frac{1}{2}$$$$g\left(x\right)=\frac{\sqrt[3]{7+x^2}-\sqrt{3+x^2}}{x-1}$$$$g(x+1)=\frac{\sqrt[3]{8+2x+x^2}-\sqrt{4+2x+x^2}}{x}$$$$g(x+1)=\frac{2\sqrt[3]{1+\frac{1}{4}x+\frac{1}{8}{x}^2}-2\sqrt{1+\frac{1}{2}x+\frac{1}{4}{x}^2}}{x}$$$$g(x+1)\underset{0}{\sim }\frac{2(1+\frac{1}{3}(\frac{1}{4}x+\frac{1}{8}{x}^2))-2(1+\frac{1}{2}(\frac{1}{2}x+\frac{1}{4}{x}^2))}{x}$$$$g(x+1)\underset{0}{\sim }\frac{\frac{2}{3}(\frac{1}{4}x+\frac{1}{8}{x}^2)-(\frac{1}{2}x+\frac{1}{4}{x}^2)}{x}\to \frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$$$$$$$$$$$$$$

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