Mit Hilfe der dritten binomischen Formel erhält man
\(\displaystyle \lim_{n\rightarrow \infty} \sqrt{2n^2+4n-7}-\sqrt{2n(n+1)}= \lim_{n\rightarrow \infty} \frac{( \sqrt{2n^2+4n-7}-\sqrt{2n(n+1)})\cdot( \sqrt{2n^2+4n-7}+\sqrt{2n(n+1)}) }{( \sqrt{2n^2+4n-7}+\sqrt{2n(n+1)})}\)
\(\displaystyle= \lim_{n\rightarrow \infty} \frac{2n^2+4n-7-2n(n+1) }{\sqrt{2n^2+4n-7}+\sqrt{2n(n+1)}}= \lim_{n\rightarrow \infty} \frac{2n^2+4n-7-2n^2-2n }{\sqrt{2n^2+4n-7}+\sqrt{2n(n+1)}}\)
\(\displaystyle=\lim_{n\rightarrow \infty} \frac{2n-7 }{\sqrt{2n^2+4n-7}+\sqrt{2n(n+1)}}=\lim_{n\rightarrow \infty} \frac{n\cdot(2-7\frac1n) }{n\cdot\sqrt{2+4\frac1n-7\frac1{n^2}}+n\cdot\sqrt{2+\frac1n}}=\frac2{2\cdot\sqrt{2}}=\frac1{\sqrt2} \)
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