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Use the definition of limit to prove that:

$$\lim_{x \to -2} \frac{x+1}{\sqrt{x^2-3}}=-1$$

so what I am trying to do is that make $\vert \frac{x+1}{\sqrt{x^2-3}} +1 \vert$ $\le$ to something. so I need to make $x+1$ big and make $\sqrt{x^2-3}$ small.

$\vert \frac{x+1}{\sqrt{x^2-3}} +1 \vert =\cdots$ some black magic here $\cdots= \frac{2|x+2|}{(\sqrt{x^2-3})[(x+1)-(\sqrt{x^2-3})]}$ and bounded we have

$0.61 < x^2-3 \lt 4.41$ and $\sqrt{0.61} < \sqrt{x^2-3} < \sqrt{0.441}$

now $-1.2 < -\sqrt{x^2-3} < -0.7$

$$-1.1 < x+1 < -0.9\text{ and }-2.1\lt x < -1.9$$

therefore $(x+2) < 1 $

$$-3 < x < -1$$

$$-2 < x^2-3 < 6$$

how would you end this prove or do i need more??..thank you for the help

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  • $\begingroup$ What is the question here? $\endgroup$
    – user142198
    Dec 7, 2014 at 5:23

2 Answers 2

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So I need to make $x+1$ big and make $\sqrt{x^2-3}$ small.

This is not correct. You need to show that given any $\epsilon > 0$ there exists $\delta > 0$ such that if $|-2-x| < \delta$, then the quantity $\left| \frac{x+1}{\sqrt{x^2-3}} +1 \right| < \epsilon$.

If you want to show something is small, bring the $1$ into the numerator:

$\left| \frac{x+1 + \sqrt{x^2-3}}{\sqrt{x^2-3}} \right|$

Show that the bottom doesn't get very small (e.g. show it's bounded below by some fixed positive number whenever $\delta < \frac{1}{2}$, say), but the top can be made arbitrarily small by choosing $\delta$ small enough.

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For a given $\epsilon > 0$, choose $\delta = \text{min}\{0.01, \frac{\epsilon}{4}\}$. So if $|x-(-2)| = |x+2| < \delta$, then: We have: $|x+1-\sqrt{x^2-3}| =\left|x+(1-\sqrt{x^2-3})\right| \geq |x| - \left|1-\sqrt{x^2-3}\right| = |x| - \dfrac{|4-x^2|}{1+\sqrt{x^2-3}} \geq |x| - |4-x^2| = |x| - |2-x||x+2| \tag{1}$.

Since $|x+2| < 0.01$, this implies that: $-2.01 < x < -1.99 \to |x| > 1.99$, and $|2-x| = 2-x < 2.01+2=4.01$. Also $\sqrt{x^2-3} \geq \sqrt{1.99^2-3} > 0.97$. Thus:

$\sqrt{x^2-3}\left|x+1-\sqrt{x^2-3}\right| \geq 0.97\left(1.99-4.01\cdot 0.01\right) > 1.8 > 1$. So:

$\dfrac{4|x+2|}{\sqrt{x^2-3}\left|x+1-\sqrt{x^2-3}\right|} < 4|x+2| < 4\cdot \dfrac{\epsilon}{4} = \epsilon$.

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