SolSept0806

$\lim_{x\rightarrow 0}{{{\sqrt{x^2+1}-\sqrt{1-x^2}}\over{1-\cos x}}}$

This is an indeterminate form because $\lim_{x\rightarrow 0}\ (1-\cos x)= 0.$

Apply l'Hôpital's rule:

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

This still is an indeterminate form because $\lim_{x\rightarrow 0} \sin x = 0.$

So apply l'Hôpital's rule once again:

$\lim_{x\rightarrow 0}{{{{{1}\over{\left(x^2+1\right)^{{{3}\over{2}} }}}+{{1}\over{\left(1-x^2\right)^{{{3}\over{2}}}}}}\over{\cos x}}}={{{{{1}\over{\left(0^2+1\right)^{{{3}\over{2}} }}}+{{1}\over{\left(1-0^2\right)^{{{3}\over{2}}}}}}\over{\cos 0}}}=2$

$\square$

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