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In a practice problem, I am supposed find the limit of the following using the squeeze theorem:

$$\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

I understand how to apply the squeeze theorem with trigonometric expressions, but I am not sure what expression I am supposed to start with using an expression like this.

Here is my attempt:

$$ \text{for } 0\le x \le3\\0 \le \sqrt{3-x} \le \sqrt{3}\\ -1 \le \sqrt{3-x}-1 \le \sqrt{3} -1\\ \frac{1}{\sqrt{3}-1} \le \frac{1}{\sqrt{3-x}-1} \le -1\\ \frac{\sqrt{6-x}-2}{\sqrt{3}-1} \le\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \le -\sqrt{6-x}+2 $$

Which then evaluates to the limit being 0, which is the incorrect answer.

elimirks
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    I think you should first rationalize the numerator or denominator before trying to apply the squeeze theorem. – Viktor Vaughn May 19 '14 at 17:47
  • Why do you want the squeeze lemma at all? Just rationalize both expressions and take the limit. I got $\frac{1}{2}$ – Alex May 19 '14 at 17:48
  • They just said to use the squeeze theorem to evaluate it. – elimirks May 19 '14 at 17:51
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    @elimirks You have the right idea, but weren't careful enough in taking reciprocals. Here's a hint: suppose that $-1 \leq x \leq 1$. Then, taking reciprocals as you did above, $1 \leq 1/x \leq -1$. Hm! – Théophile May 19 '14 at 18:15
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    I wouldn't use the squeeze theorem at all on this. – Michael Hardy May 19 '14 at 18:20

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