Question

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$lim_{x to 0} frac{ sqrt{1+ x^{2} }-1 }{ sqrt{ x^{2} +16} -4}$

Answer 1

$lim_{x to 0} frac{ sqrt{1+ x^{2} }-1 }{ sqrt{ x^{2} +16} -4} =frac{0}{0}= lim_{x to 0} frac{ sqrt{1+ x^{2} }-1 }{ sqrt{ x^{2} +16} -4}*frac{ sqrt{1+ x^{2} }+1}{ sqrt{1+ x^{2} }+1}*frac{sqrt{ x^{2} +16} +4}{sqrt{ x^{2} +16} +4}=\= lim_{x to 0}frac{x^2*(sqrt{x^2+16}+4)}{x^2(sqrt {1+x^2}+1}= lim_{x to 0}frac{sqrt{x^2+16}+4}{sqrt {1+x^2}+1}=4$