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Prove $$\lim_{x\to0}\sqrt{4-x}=2$$ using the precise definition of limits. (Epsilon-Delta)

I am not sure how to link $0<\left |x \right |<\delta$ with $\left |\sqrt{4-x}-2\right |<\epsilon$ .

EDIT (Trying it out now)

I worked till here, then I basically got stuck.

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2 Answers 2

up vote 2 down vote accepted

First, you can try and prove some important yet simple facts about limits. The following are all equivalent:

$$\eqalign{ & \mathop {\lim }\limits_{x \to a} f\left( x \right) = l \cr & \mathop {\lim }\limits_{h \to 0} f\left( {a + h} \right) = l \cr & \mathop {\lim }\limits_{h \to 0} f\left( {a- h} \right) = l \cr & \mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) - l} \right] = 0 \cr} $$

You have

$$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$

or

$$\mathop {\lim }\limits_{h \to 0} \sqrt {4 - h} + 2 = 4$$

By the above, this is equivalent to

$$\mathop {\lim }\limits_{x \to 4} \sqrt x + 2 = 4$$

So you want to show that for each $\epsilon >0$ there is a $\delta>0$ such that for all $x$, $$0 < \left| {x - 4} \right| < \delta \Rightarrow \left| {\sqrt x - 2} \right| < \varepsilon $$

But multiplying the conjugate gives $$\left| {\sqrt x - 2} \right| = \left| {\frac{{x - 4}}{{\sqrt x + 2}}} \right| < \left| {x - 4} \right|$$

so taking $\delta=\epsilon$ does it.

NOTE You could've also worked with $$\mathop {\lim }\limits_{x \to 0} \sqrt {4 - x} + 2 = 4$$

in fact,

$$\left| {\sqrt {4 - x} + 2 - 4} \right| = \left| {\sqrt {4 - x} - 2} \right| = \left| {\frac{{ - x}}{{\sqrt {4 - x} + 2}}} \right| < \left| x \right|$$

so again, $\delta=\epsilon$, as expected.

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Based on your correction, I recommend the following approach.

If we set $y:=4-x$ (to clean up the expression), then $x\to 0$ precisely as $y\to 4$, yes? Thus, we may equivalently show that $$\lim_{y\to 4}\sqrt{y}=2.\tag{1}$$ Fixing some $\varepsilon>0$, we must find $\delta>0$ such that for $0<|y-4|<\delta$, we have $|\sqrt{y}-2|<\varepsilon$. It's worth noting here that $$|y-4|=|(\sqrt{y}+2)(\sqrt{y}-2)|=|\sqrt{y}+2|\cdot|\sqrt{y}-2|\tag{2}$$ wherever each expression is defined. Thus, noting that $\sqrt{y}+2$ is positive for positive real $y$, we have $$\frac1{\sqrt{y}+2}\leq\frac12<1\tag{3}$$ for positive real $y$. Hence, for any $\delta>0$, we have by $(2)$ and $(3)$ that $$0<|y-4|<\delta\quad\Rightarrow\quad|\sqrt{y}-2|<\frac1{\sqrt{y}+2}\delta<\delta,\tag{4}$$ wherever each expression is defined.

Now, it looks like we can just set $\delta=\varepsilon$ and be done with it, yes? There's only one potential problem: What if one of the expressions is undefined? Indeed, if our arbitrary $\varepsilon>0$ is too large, we may well be making claims involving square roots of negative numbers--for example, if we had chosen $\varepsilon=5$, then $y=-1/2$ satisfies $|y-4|<\varepsilon$, but $|\sqrt{y}-2|$ doesn't make any sense (in this context). Thus, to ensure that $y\geq 0$, we need $|y-4|\leq 4$, so we set $\delta:=\min\{4,\varepsilon\},$ whence each expression in $(4)$ is defined, and $$0<|y-4|<\delta\quad\Rightarrow\quad|\sqrt{y}-2|<\delta\leq\varepsilon,\tag{5}$$ as desired.

To get the result in terms of $x$, we need only substitute $y=4-x$ in each instance, and observe that $|y-4|=|-x|=|x|=|x-0|$, so everything looks just as it should, and shows that the two approaches (using $x$ vs. using $y=4-x$) are indeed equivalent.

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Thanks for pointing that out. I adjusted the question –  Yellow Skies Sep 24 '12 at 3:57
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