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_{n}such that [itex] \frac{{m_n}^2}{n^2} < 2 < \frac{{(m_n + 1)^2}}{n^2} [/itex]. show that [itex] {q_n}^2 [/itex]approaches 2 as n approaches infinity.

i need to show that [itex] |\frac{{m_n}^2}{n^2} - 2| < ε [/itex]. what i have so far is that [itex] 2 - \frac{{m_n}^2}{n^2} < \frac{{(m_n + 1)^2}}{n^2} - \frac{{m_n}^2}{n^2} = \frac{2m_n + 1}{n^2} [/itex].

my trouble is that i can't figure out a way to bound m_n. i know how to take care of the n

^{2}by choosing N such that N

^{2}ε > 1 and choosing n > N. can someone give me a hint on how to deal with the m_n? thanks.