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$$\lim_{n\rightarrow \infty} \frac{\sqrt{n+a}-\sqrt{n+b}}{\sqrt{n+c}-\sqrt{n+d}},\textrm{ } (c\neq d)$$ I really don't know what to use to solve this. Any help would be greatly appreciated.


Following your sugestions I get

$$\lim_{n\rightarrow \infty} \frac{\sqrt{n+a}-\sqrt{n+b}}{\sqrt{n+c}-\sqrt{n+d}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n+a}+\sqrt{n+b}}{\sqrt{n+c}+\sqrt{n+d}}\left(\frac{a-b}{c-d}\right)=\left(\frac{a-b}{c-d}\right)$$

That should be it, thanks.

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    $\begingroup$ Hint: Multiply numerator and denominator with $\sqrt{n+c}+\sqrt{n+d}$ $\endgroup$ – gammatester Nov 17 '16 at 15:16
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    $\begingroup$ Another idea is to try binomial expansion for a few terms. $\endgroup$ – Simply Beautiful Art Nov 17 '16 at 15:18
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    $\begingroup$ Also, multiply and divide by $\sqrt{n+a}+\sqrt{n+b}$. $\endgroup$ – Galc127 Nov 17 '16 at 15:18
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    $\begingroup$ @Galc127 I presume you mean do this in addition to what gammatester suggests - carefully done that helps a lot. $\endgroup$ – Mark Bennet Nov 17 '16 at 15:27
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    $\begingroup$ @MarkBennet, of course. We need to do both in order to evaluate the limit. $\endgroup$ – Galc127 Nov 17 '16 at 15:28
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After factoring out by $\sqrt{n}$ and simplifying,

and with

$$\sqrt{1+\frac{a}{n}}=1+\frac{a}{2n}(1+\epsilon(n)),$$

we get the limit

$$\frac{a-b}{c-d}$$

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