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Question Number 99936 by student work last updated on 24/Jun/20

$$\mathrm{li}\underset{\mathrm{x}\to 0}{\mathrm{m}}\frac{{(1+\mathrm{x})}^{\mathrm{k}}-1}{\mathrm{x}}=?$$$$\mathrm{help}\mathrm{me}$$

Commented by bemath last updated on 24/Jun/20

$$\mathrm{for}\mathrm{k}>0\Rightarrow \underset{x\to 0}{\lim }\frac{\mathrm{k}{(1+\mathrm{x})}^{\mathrm{k}-1}.1}{1}=\mathrm{k}$$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

$$\underset{x\to 0}{\lim }\frac{{(1+x)}^k-1}{1+x-1}=k.1=k$$

Commented by Dwaipayan Shikari last updated on 24/Jun/20

$$k$$

Answered by mathmax by abdo last updated on 24/Jun/20

$$\mathrm{take}\mathrm{a}\mathrm{try}\mathrm{first}\mathrm{and}\mathrm{dont}\mathrm{waste}\mathrm{your}\mathrm{time}...$$

Answered by mathmax by abdo last updated on 24/Jun/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{{(1+\mathrm{x})}^{\mathrm{n}}-1}{\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\sum _{\mathrm{k}=0}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}-1}{\mathrm{x}}=\frac{\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}{\mathrm{x}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}-1}{\mathrm{C}}_{\mathrm{n}}^{\mathrm{k}+1}{\mathrm{x}}^{\mathrm{k}}={\mathrm{C}}_{\mathrm{n}}^1+{\mathrm{C}}_{\mathrm{n}}^2\mathrm{x}+{\mathrm{C}}_{\mathrm{n}}^3{\mathrm{x}}^2+...+{\mathrm{C}}_{\mathrm{n}}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow {\lim }_{\mathrm{x}\to 0}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{C}}_{\mathrm{n}}^1=\mathrm{n}$$

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