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I would like some help with the following problem. Thanks for any help in advance.

Let $(x_n)$ and $(y_n)$ be convergent sequences of positive real numbers. Let $ x_n \xrightarrow[n \to \infty]{} x$ and $ y_n \xrightarrow[n \to \infty]{} y$ and suppose that $x > 0$. Prove that the sequence $(x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2$, $n \geq 1$ converges.

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    $\begingroup$ The universal procedure, multiply the top and missing bottom by $(x_n n^4+y_n^2)^{1/2} +x_n^{1/2}n^2$. $\endgroup$ – André Nicolas Feb 13 '16 at 21:23
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Expanding on André Nicolas's comment,

$\begin{array}\\ (x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2 &=((x_nn^4 + y_nn^2)^{1/2} - x_n^{1/2}n^2) \dfrac{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{(x_nn^4 + y_nn^2)- x_nn^4}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_nn^2}{(x_nn^4 + y_nn^2)^{1/2} + x_n^{1/2}n^2}\\ &=\dfrac{ y_n}{(x_n + y_n/n^2)^{1/2} + x_n^{1/2}}\\ \end{array} $

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