# How can I solve this graph using the precise definition of limits?

Use the given graph of $f(x)=\sqrt{x}$ to find a number $\delta$ such that

if $|x-4|<\delta$ then $|\sqrt{x}-2|<0.4$

This precise definition of a limit has been giving me a lot of trouble, but so far I have the left value of x $f(x)=1.6$ $\sqrt{x}=1.6$ $x=2.56$, and the right value $f(x)=2.4$ $\sqrt{x}=2.4$ $x=5.76$

Now, I am stuck because I don't know what to do with this information. I think my main problem is that I don't understand this concept in general. I've been going through my textbook and the problems, I still don't get it. So far I've been able to do questions that already have all of the values of $x$ and $f(x)$ on the graph, but only because I have remembered what steps I'm supposed to take, not that I understand what I'm doing (not even the tiniest bit).

What are my next steps?

## 2 Answers

Take $\delta=4-2.56$, so each element in the interval $]4-\delta,4+\delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.

• Am I trying to find a number $\delta$ that is close to 4, so the limit is brought close to a $f(x)$ value that is less than 0.4? I apologize in advance is this question doesn't make much sense, this concept has been really hard for me to wrap my head around. Jun 16, 2015 at 18:11
• you are searching for $\delta$ to make a $\delta$-neighbourhood of $4$ such that for each point in it, they are carried $0.4$-near to $2$. Jun 16, 2015 at 18:16

$$|\sqrt{x}- 2|< 0.4$$ is the same as $$-0.4< \sqrt{x}- 2< 0.4$$

Since $$\sqrt{x}$$ is an increasing function, it is sufficient to look at the endpoints of the interval, $$-0.4= \sqrt{x}- 2$$ and $$\sqrt{x}- 2|= 0.4$$.

For $$-0.4= \sqrt{x}- 2$$ we have immediately $$\sqrt{x}= -0.4+ 2= 1.6$$. Squaring both sides, $$x= 2.56$$. For $$\sqrt{x}- 2= 0.4$$, $$\sqrt{x}= 2+ 0.4= 2.4$$ and $$x= 5.76$$.

That is, $$|\sqrt{x}- 4|< 0.4$$ is true as long as $$2.56< x< 5.76$$.

$$2.56= 4- 1.44$$ and $$5.76= 4+ 1.74$$. The smaller of those is 1.44 so $$|x- 4|< 1.44$$ will fit both.