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Решите задачу:

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