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$\displaystyle{}\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$ is given by

  1. $0$
  2. $-1$
  3. $1$
  4. $\frac{1}{2}$
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$\lim_{x->0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$  [putting x=0, it comes 0/0 form , So, apply L hospital rule(differentiate upper limit and lower limit differently)]

$\lim _{x \rightarrow 0}\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}=\frac{1}{2}+\frac{1}{2}=1$

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