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Show the following limit, using the definition of epsilon-delta. $$\lim_{x\rightarrow\sqrt{2}}\frac{\big[|x|\big]+x}{3+x-x^{2}}=1$$ I have problems with the floor function, I do not know if I should work in the numerator like: $ 1 + x $ directly or I'm doing it wrong.

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As $x$ is confined to a sufficiently small neighborhood of $\sqrt{2}$, what can we say about the sign of $x$? And what then would be $|x|$? And what would be $\lfloor | x | \rfloor$?

If the above reasoning lacks sufficient rigor, compute a radius $\rho > 0$ around $\sqrt{2}$ for which, whenever $|x - \sqrt{2}| < \rho$, we would have $1 \le |x| < 2$, consequently $\lfloor |x| \rfloor = 1$. Then when $\delta < \rho$, we can consider instead the behavior of the function $$\frac{1+x}{3+x-x^2}$$ in a neighborhood of $\sqrt{2}$ with radius $\delta$.

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I will provide a hint:  ¿Can you find a $\delta$ such that given any (small) $\epsilon$, say $\epsilon=1$ (not particularly small...but just perfect here) $$\left|\left\lfloor x\right\rfloor - 1\right| < 1 = \epsilon$$ for all $\left|x-\sqrt{2}\right| < \delta$.

Second hint:  $0 < 1$...and by definition $0 < \epsilon$.

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  • $\begingroup$ I have a doubt, because I discard the denominator or can I find a dimension for that? ... and thanks $\endgroup$ – user551505 Apr 12 '18 at 1:36

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