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.


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

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


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

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.


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.

  • 1
    $\begingroup$ Thanks for pointing that out. I adjusted the question $\endgroup$ – Yellow Skies Sep 24 '12 at 3:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.