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?
Answer
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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