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finn grensa under:

[tex]\lim_{n \to \infty}\left(\frac{1}{\sqrt{n}\sqrt{n+1}}\,+\,\frac{1}{\sqrt{n}\sqrt{n+2}}\,+\,...\,+\,\frac{1}{\sqrt{n}\sqrt{n+n}} \right )[/tex]

[tex]\lim_{n\rightarrow \infty}\left ( \frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+ \dots +\frac{1}{\sqrt{n}\sqrt{n+n}} \right )=\lim_{n\rightarrow \infty}\left ( \sum_{i=1}^{n}\frac{1}{\sqrt{n}\sqrt{n+i}} \right )= \lim_{n\rightarrow \infty}\left ( \sum_{i=1}^n \frac{1}{n}\frac{1}{\sqrt{1+\frac{i}{n}}} \right ) =\int_{0}^1\frac{1}{\sqrt{1+x}}dx[/tex]


La [tex]\frac{1}{\sqrt{1+x}}=f(x)[/tex]


[tex]u=1+x[/tex]
[tex]du=1[/tex]

[tex]\int\frac{du}{\sqrt{u}}=2\sqrt{u}+C=2\sqrt{x+1}+C[/tex]


[tex]\int_0^1f(x)dx=\left [ 2\sqrt{1+x} \right ]_0^1=\left [ 2\sqrt{1+1} \right ]-\left [ 2\sqrt{1+0} \right ]=2\sqrt{2}-2=2(\sqrt{2}-1)[/tex]

Alt i alt [tex]\lim_{n\rightarrow \infty}\left ( \frac{1}{\sqrt{n}\sqrt{n+1}}+\frac{1}{\sqrt{n}\sqrt{n+2}}+ \dots +\frac{1}{\sqrt{n}\sqrt{n+n}} \right )=2(\sqrt{2}-1)[/tex]

Ikke helt triviell oppgave, fin løsning.