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$$\lim_{x\rightarrow 0}\frac{\sqrt{1+2x}-\sqrt{1-2x}}{\sin x}$$

What I did was use binomial theorem and the fact that $\lim_{x\rightarrow 0}\dfrac{\sin x}{x} = 1$.

$$\lim_{x \rightarrow 0}\frac{2\left[x+{1/2\choose{3}}8x^3+{1/2\choose{5}}32x^5...\right]}{\sin x} = 2 $$

Can we do this by any more simpler way, by using simple factorisation? Thank you!

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Hint: use the $\sqrt{1+2x}-\sqrt{1-2x}={({\sqrt{1+2x}-\sqrt{1-2x})(\sqrt{1+2x}+\sqrt{1-2x})}\over{\sqrt{1+2x}+\sqrt{1-2x}}}$

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  • $\begingroup$ Thank you! I think i need to revisit elementary algebra :| $\endgroup$ – Max Payne Jan 21 '16 at 4:43
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$$\lim_{x\rightarrow 0}\frac{\sqrt{1+2x}-\sqrt{1-2x}}{\sin x}=\lim_{x\to 0}\frac{1+2x-1+2x}{\sin x(\sqrt{1+2x}+\sqrt{1-2x})}$$ $$=(\lim_{x\to 0}\frac{1}{\sqrt{1+2x}+\sqrt{1-2x}})\cdot(\lim_{x\to 0}\frac{4x}{\sin x})=\frac{1}{2}\cdot 4=2$$

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Use $(a-b)(a+b) = a^2-b^2$ to expand the fraction, then divide by $x$:

$$\lim_{x\rightarrow 0}\frac{\sqrt{1+2x}-\sqrt{1-2x}}{\sin x} = \lim_{x\rightarrow 0}\frac{(1+2x)-(1-2x)}{\sin x(\sqrt{1+2x}+\sqrt{1-2x})} = \lim_{x\rightarrow 0}\frac{4x}{\sin x(\sqrt{1+2x}+\sqrt{1-2x})} = \lim_{x\rightarrow 0}\frac{4}{{\sin x\over x}(\sqrt{1+2x}+\sqrt{1-2x})} = {4 \over 1\cdot 2} = 2$$

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$\begin{eqnarray}\frac{\sqrt{1+2x}-\sqrt{1-2x}}{\sin x}&=&\frac{\sqrt{1+2x}-\sqrt{1-2x}}{\sin x}\times\frac{\sqrt{1+2x}+\sqrt{1-2x}}{\sqrt{1+2x}+\sqrt{1-2x}}\\&=&\frac{4x}{\sin x(\sqrt{1+2x}+\sqrt{1-2x})}\\ &=&4\frac{x}{\sin x}\frac{1}{\sqrt{1+2x}+\sqrt{1-2x}}\\&\to&4\times 1\times\frac{1}{2}\\ &=&2\end{eqnarray}$

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