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If $x_n$ $\rightarrow $ 1

Then show that sequence $\frac {4+ (x_n)^{2}}{x_n}$ approaches to limit 5

I have tried to find epsilon proof ,But i am not successful .Can anyone help me with this

Thanks

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First you can prove that $ \dfrac{1}{2} < x_n < 2$ is true for some $n \geq N$. And use this to show that: $\left|\dfrac{4+x_n^2}{x_n}-5\right| = \left|\dfrac{x_n^2-5x_n+4}{x_n}\right| = \left|x_n-1\right|\cdot \dfrac{\left|x_n-4\right|}{|x_n|}\leq \left|x_n-1\right|\cdot \dfrac{\left(|x_n|+4\right)}{\frac{1}{2}}\leq \left|x_n-1\right|\cdot 2(2+4)= 12|x_n-1|, n \geq N$ and the $\epsilon-\delta$ argument should work.

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  • $\begingroup$ You are welcome, buddy..... $\endgroup$
    – DeepSea
    May 4 '15 at 2:11
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How about a direct computation $$ \lim_{n \to \infty} \frac{4+(x_n)^2}{x_n} = \lim_{n \to \infty} \frac{4}{x_n} + \lim_{n \to \infty} \frac{(x_n)^2}{x_n} = \frac41 + \lim_{n \to \infty} x_n = 4+1 = 5 $$

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