Граница на функция.. +безкрайност

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[tex]\lim_{x \to \infty}[/tex] [tex](\sqrt{n^{2}-n+3}-{\sqrt{n^{2}+1}})[/tex]

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[tex]\lim_{x \to \infty}(\sqrt{ n^2-n+3}-\sqrt{ n^2+1})=[/tex]

[tex]=\lim_{n \to \infty}\frac{(\sqrt{ n^2-n+3}-\sqrt{ n^2+1})(\sqrt{ n^2-n+3}+\sqrt{ n^2+1})}{\sqrt{ n^2-n+3}+\sqrt{ n^2+1}}=[/tex]

[tex]=\lim_{n \to \infty}\frac{(\sqrt{ n^2-n+3})^2-(\sqrt{ n^2+1})^2}{\sqrt{ n^2-n+3}+\sqrt{ n^2+1}}=[/tex]

[tex]=\lim_{n \to \infty} \frac {n^2-n+3-( n^2+1)}{\sqrt { n^2-n+3}+\sqrt{ n^2+1}}=[/tex]

[tex]=\lim_{n \to \infty}\frac{-n+2}{n \left ( \sqrt { 1-\frac {1}{n}+\frac {3}{n^2}}+\sqrt{ 1+\frac{1} {n^2}} \right )}=[/tex]

[tex]=-\lim_{n \to \infty}\frac{\cancel {n} \left ( 1+\frac{2}{n} \right )}{\cancel {n} \left ( \sqrt { 1-\frac {1}{n}+\frac {3}{n^2}}+\sqrt{ 1+\frac{1} {n^2}} \right )}=- \frac{1+0}{\sqrt{1-0+3.0^2}+\sqrt{1+2.0}}=-\frac{1}{2}[/tex]

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