Question #15547

Using the "E-N" definition of the limit, show that

lim n[(n^2+2)^(1/2) - n] = 1

Expert's answer

limnn[n2+2n]=limnn(n2+2n)(n2+2+n)n2+1+n=limnn(n2+2n2)n2+2+n==limn2nn2+2+n=limn21/n2+2+1=22+1\begin{array}{l} \lim _ {n \to \infty} n \Big [ \sqrt {n ^ {2} + 2} - n \Big ] = \lim _ {n \to \infty} \frac {n \Big (\sqrt {n ^ {2} + 2} - n \Big) \Big (\sqrt {n ^ {2} + 2} + n \Big)}{\sqrt {n ^ {2} + 1} + n} = \lim _ {n \to \infty} \frac {n \Big (n ^ {2} + 2 - n ^ {2} \Big)}{\sqrt {n ^ {2} + 2} + n} = \\ = \lim _ {n \to \infty} \frac {2 n}{\sqrt {n ^ {2} + 2} + n} = \lim _ {n \to \infty} \frac {2}{\sqrt {1 / n ^ {2} + 2} + 1} = \frac {2}{\sqrt {2} + 1} \end{array}

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