Using the "E-N" definition of the limit, show that
lim n[(n^2+2)^(1/2) - n] = 1
lim n → ∞ n [ n 2 + 2 − n ] = lim n → ∞ n ( n 2 + 2 − n ) ( n 2 + 2 + n ) n 2 + 1 + n = lim n → ∞ n ( n 2 + 2 − n 2 ) n 2 + 2 + n = = lim n → ∞ 2 n n 2 + 2 + n = lim n → ∞ 2 1 / n 2 + 2 + 1 = 2 2 + 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} lim n → ∞ n [ n 2 + 2 − n ] = lim n → ∞ n 2 + 1 + n n ( n 2 + 2 − n ) ( n 2 + 2 + n ) = lim n → ∞ n 2 + 2 + n n ( n 2 + 2 − n 2 ) = = lim n → ∞ n 2 + 2 + n 2 n = lim n → ∞ 1/ n 2 + 2 + 1 2 = 2 + 1 2
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