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Question Number 131882 by Eric002 last updated on 09/Feb/21

$$\underset{x\to 0}{\lim }\frac{x}{x+\sqrt[4]{x+1}-1}$$

Answered by liberty last updated on 09/Feb/21

$${\mathrm{L}}^{\prime }\mathrm{H}\ddot{\mathrm{o}pital}\mathrm{L}=\underset{x\to 0}{\lim }\left[\frac{1}{1+\frac{1}{4\sqrt[4]{{(\mathrm{x}+1)}^3}}}\right]=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$$$$\mathrm{or}\underset{x\to 0}{\lim }\left[\frac{\mathrm{x}}{\mathrm{x}+1+\frac{\mathrm{x}}{4}-1}\right]=\underset{x\to 0}{\lim }\frac{4\mathrm{x}}{5\mathrm{x}}=\frac{4}{5}$$

Commented by Eric002 last updated on 09/Feb/21

$$howdidyouget\frac{x}{x+1+\frac{x}{4}-1}$$

Commented by EDWIN88 last updated on 09/Feb/21

$$\mathrm{Bernoulli}\mathrm{equation}$$$$\underset{x\to 0}{\lim }{(1+\mathrm{f}\left(\mathrm{x}\right))}^{\mathrm{n}}\approx \underset{x\to 0}{\lim }(1+\mathrm{n}.\mathrm{f}\left(\mathrm{x}\right))$$$$\mathrm{so}\underset{x\to 0}{\lim }\sqrt[4]{1+\mathrm{x}}\approx \underset{x\to 0}{\lim }(1+\frac{\mathrm{x}}{4})$$

Commented by Eric002 last updated on 09/Feb/21

$$thankyousir$$

Answered by EDWIN88 last updated on 09/Feb/21

$$\mathrm{the}\mathrm{another}\mathrm{way}$$$$\mathrm{let}\sqrt[4]{\mathrm{x}+1}=\ell \mathrm{where}\ell \to 1\mathrm{and}\mathrm{x}={\ell }^4-1$$$$\mathrm{L}=\underset{\ell \to 1}{\lim }\frac{{\ell }^4-1}{{\ell }^4+\ell -2}=\underset{\ell \to 1}{\lim }\frac{({\ell }^2+1)(\ell +1)(\ell -1)}{(\ell -1)({\ell }^3+{\ell }^2+\ell +2)}$$$$=\underset{\ell \to 1}{\lim }\frac{({\ell }^2+1)(\ell +1)}{{\ell }^3+{\ell }^2+\ell +2}=\frac{4}{5}$$$$$$

Answered by mathmax by abdo last updated on 10/Feb/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}{+}^4\sqrt{1+\mathrm{x}}-1}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\mathrm{x}+{(1+\mathrm{x})}^{\frac{1}{4}}-1}$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)\sim \frac{\mathrm{x}}{\mathrm{x}+1+\frac{\mathrm{x}}{4}-1}=\frac{\mathrm{x}}{\frac{5}{4}\mathrm{x}}=\frac{4}{5}\Rightarrow {\lim }_{\mathrm{x}⌣)0}\mathrm{f}\left(\mathrm{x}\right)=\frac{4}{5}$$

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