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I'm stuck trying to find the limit of the sequence $$\frac{\sqrt{12 + a_n} - \sqrt{4a_n}}{a_n^2 - 2a_n - 8}$$

Where I'm given that $a_n > 4$ and $a_n \rightarrow 4$

Both the numerator and the denominator tend to 0, and I can't find how to solve this indetermination. I tried multiplying and dividing by the "reciprocal" of the numerator to get rid of the square roots in the numerator, but that doesn't seem to lead anywhere. What else can I try?

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  • $\begingroup$ Rationalize the numerator $\endgroup$ – lab bhattacharjee Nov 21 '19 at 22:39
  • $\begingroup$ that's what I did, but it's still an indetermination $\endgroup$ – Francisco José Letterio Nov 21 '19 at 22:41
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Hint:

$$b^2-2b-8=(b-4)(b+2)$$

$$\sqrt{12+b}-\sqrt{4b}=-\dfrac{3(b-4)}{\sqrt{12+b}+\sqrt{4b}}$$

If $b\to4,b\ne4\implies b-4\ne0$ hence can be cancelled safely

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We can use the strandar trick

$$\frac{\sqrt{12 + a_n} - \sqrt{4a_n}}{a_n^2 - 2a_n - 8}=\frac{\sqrt{12 + a_n} - \sqrt{4a_n}}{a_n^2 - 2a_n - 8}\cdot \frac{\sqrt{12 + a_n} + \sqrt{4a_n}}{\sqrt{12 + a_n} + \sqrt{4a_n}}=$$

$$=\frac{3\color{red}{(4 - a_n)}}{\color{red}{(a_n-4)}(a_n+2)}\frac{1}{\sqrt{12 + a_n} + \sqrt{4a_n}}$$

to eliminate the indetermination.

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