Question

Знайти границі послідовності.

Answer 1

Ответ:

$\displaystyle 4$

Объяснение:

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

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

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

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

$\displaystyle \lim_{x\to 0}\frac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\frac{\sqrt{0^2+16}+4}{\sqrt{0^2+1}+1}=\\\\\\\frac{\sqrt{0+16}+4}{\sqrt{0+1}+1}=\frac{\sqrt{16}+4}{\sqrt{1}+1}=\frac{4+4}{1+1}=\frac{8}{2}=4$