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Question Number 161437 by cortano last updated on 17/Dec/21

$$\underset{x\to 0}{\lim }\frac{\sqrt[10]{8x^2-4x+1}+\sqrt[5]{7x^2+3x+1}-2}{x}=?$$

Answered by bobhans last updated on 18/Dec/21

$$\underset{x\to 0}{\lim }\frac{\sqrt[10]{{8\mathrm{x}}^2-4\mathrm{x}+1}-1+\sqrt[5]{{7\mathrm{x}}^2+3\mathrm{x}+1}-1}{\mathrm{x}}$$$${\mathrm{L}}_1=\underset{x\to 0}{\lim }\frac{(8\mathrm{x}-4)[\sqrt[10]{{8\mathrm{x}}^2-4\mathrm{x}+1}-1]}{({8\mathrm{x}}^2-4\mathrm{x}+1)-1}$$$${\mathrm{L}}_1=-4\times \underset{\mathrm{r}\to 1}{\lim }\frac{\mathrm{r}-1}{{\mathrm{r}}^{10}-1}=-\frac{4}{10};[\mathrm{r}=\sqrt[10]{{8\mathrm{x}}^2-4\mathrm{x}+1}]$$$${\mathrm{L}}_2=\underset{x\to 0}{\lim }\frac{(7\mathrm{x}+3)[\sqrt[5]{{7\mathrm{x}}^2+3\mathrm{x}+1}-1]}{({7\mathrm{x}}^2+3\mathrm{x}+1)-1}$$$${\mathrm{L}}_2=3\times \underset{x\to 0}{\lim }\frac{\sqrt[5]{{7\mathrm{x}}^2+3\mathrm{x}+1}-1}{({7\mathrm{x}}^2+3\mathrm{x}+1)-1}$$$${\mathrm{L}}_2=3\times \underset{\mathrm{z}\to 1}{\lim }\frac{\mathrm{z}-1}{{\mathrm{z}}^5-1}=\frac{3}{5};[\mathrm{z}=\sqrt[5]{{7\mathrm{x}}^2+3\mathrm{x}+1}]$$$$\therefore \mathrm{L}={\mathrm{L}}_1+{\mathrm{L}}_2=-\frac{4}{10}+\frac{3}{5}=\frac{2}{10}=\frac{1}{5}$$

Answered by mathmax by abdo last updated on 18/Dec/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{(1+{8\mathrm{x}}^2-4\mathrm{x})}^{\frac{1}{10}}+{(1+{7\mathrm{x}}^2+3\mathrm{x})}^{\frac{1}{5}}-2}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{1+\frac{1}{10}({8\mathrm{x}}^2-4\mathrm{x})+1+\frac{1}{5}({7\mathrm{x}}^2+3\mathrm{x})-2}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\frac{4}{5}{\mathrm{x}}^2-\frac{2}{5}\mathrm{x}+\frac{7}{5}{\mathrm{x}}^2+\frac{3}{5}\mathrm{x}}{\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{\frac{11}{5}{\mathrm{x}}^2+\frac{1}{5}\mathrm{x}}{\mathrm{x}}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{11}{5}\mathrm{x}+\frac{1}{5}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{5}$$$$$$

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