Obliczyć granice

[osada]

\(\displaystyle{ a_{n}=\frac{1}\sqrt{n^2 +7n}-\sqrt{n}}\)

Pierwiastek na dole jest na d n^2 +7n nie wiem czemu wydzedł mi taki dziwny

[natkoza]

\(\displaystyle{ \lim_{n\to\infty}\frac{1}{\sqrt{n^{2}+7n}-\sqrt{n} }=\lim_{n\to\infty}\frac{1}{\sqrt{n^{2}+7n}-\sqrt{n}}\cdot \frac{\sqrt{n^{2}+7n}+\sqrt{n}}{\sqrt{n^{2}+7n}+\sqrt{n} }=\lim_{n\to\infty}\frac{\sqrt{n^{2}+n}+\sqrt{n}}{n^{2}+7n-n}=\lim_{n\to\infty}\frac{\sqrt{n^{2}(1+\frac{1}{n})}+\sqrt{n^{2}\frac{1}{n}}}{n^{2}(1+\frac{6}{n}}=\lim_{n\to\infty} \frac{n\sqrt{1+\frac{1}{n}}+n\sqrt{\frac{1}{n}}}{n^{2}(1+\frac{6}{n}}=\lim_{n\to\infty} \frac{n(\sqrt{1+\frac{1}{n}}+\sqrt{\frac{1}{n}}}{n^{2}(1+\frac{6}{n}}=\lim_{n\to\infty} \frac{\sqrt{1+\frac{1}{n}}+\sqrt{\frac{1}{n}}}{n(1+\frac{6}{n}}=0}\)