Prove 2 Limits Questions with Definition 2.4.1

[uchihajeff]

Homework Statement

Use Definition 2.4.1 to prove that the stated limit is correct.

Definition 2.4.1 in my book is:

lim as x->a of f(x) = L

if given any number e(epsilon)>0 we can find a number d(delta)>0 such that

|f(x)-L|<e if 0<|x-a|<d​

Homework Equations

Question 31. lim as x>-2 of 1/(x+1) = -1
Question 33. lim as x>4 of sqrt(x) = 2

The Attempt at a Solution

31. |1/(x+1) + 1|<e, 0<|x+2|<d
|(x+2)/(x+1)|<e
set d<=1
-1<x+2<1, -2<x+1<0
|x+1|<0
|x+2|< e * |x+1|
...then I get stuck

32. |sqrt(x)-2|<e, 0<x-4<d
sqrt(x)<e-2
x<(e-2)^2
x-4<(e-2)^2-4
...by here I'm probably already wrong
d=(e-2)^2-4

4. The answers in the back of the book
31) d=min(1,e/(1+e))
33) d=2e

P.S. Sorry, I don't know how to use Latex or whatever mathematical typing system you guys use here, so it's a little messy/unreadable.
Thanks in advance for the help!

 

[pardesi]

well one of the best way and probably only way to prove limits using the absic definition is to do like thsi

suppose i need to prove

\lim _{x/to/a} f(x)= b
that means i have to prove for every positive real number \epsilon there exists a positive delta such that

|f(x)-b| &lt; \epsilon for all real x satisfying |x-a| &lt; \delta

for a moment assume that there exists a x such that
|f(x)-b| &lt; \epsilon
manipulate such that u reach a stage where k|x-a| &lt; \epsilon
for some constant k then ur \delta is
k\epsilon
here the best way i would prefer is u looking into the proof of the fact that
\lim_{x \to a} \frac{f}{g}=\frac{\lim_{ x \to a} f}{\lim_{x \to a} g}...
and similarly for the second part
\lim_{x \to a} f^{\frac{1}{n}}={\lim f }^{\frac{1}{n}}...

 

[VietDao29]

-1<x+2<1, -2<x+1<0
|x+1|<0
|x+2|< e * |x+1|

This part is wrong, you are taking the absolute value of x + 1, and you have -2 < x + 1 < 0, you should have 0 < |x + 1| < 2, not |x + 1| < 0

...
|f(x)-b| &lt; \epsilon for all real x satisfying |x-a| &lt; \delta

Well, nope, x cannot be a, you should add '0 <' part in front of it, like this:
0 &lt; |x - a| &lt; \delta \Rightarrow |f(x) - b| &lt; \epsilon. :)

Yup, as pardesi has pointed out, you should re-arrange the expression, so that you could obtain:
|f(x) - b| < ... < k |x - a| < epsilon (where k is any constant, this is the tricky part, you should find the appropriate k)
~~~> |x - a| < epsilon / k

So we can choose delta to be (epsilon / k), so, for:
x : 0 &lt; |x - a| &lt; \delta = \frac{\epsilon}{k} \Rightarrow |f(x) - b| &lt; ... &lt; k |x - a| &lt; k \epsilon &lt; k \times \frac{\epsilon}{k} = \epsilon as required.

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I'll help you do the first problem, and give you some hints for the seconds one, they are pretty much the same. :)

Problem 1:
Prove:
\lim_{x \rightarrow -2} \frac{1}{x + 1} = -1

So, we'll start here:
\left| \frac{1}{x + 1} + 1 \right| = \left| \frac{x + 2}{x + 1} \right| = \textcolor{blue}{\left| \frac{1}{x + 1} \right|} \textcolor{red}{|x + 2|}.

Now, we will obtain the k from the blue part. We will find out the maximum value for |1 / (x+1)|, and assign k to be that value.

To find the maximum value for |1 / (x+1)|, we'll make the restriction:
\textcolor{green} {0 &lt; |x + 2| &lt; \delta \leq 0.5} \Rightarrow 0 &lt; |x + 2| &lt; 0.5 \Rightarrow -0.5 &lt; x + 2 &lt; 0.5

\Rightarrow -1.5 &lt; x + 1 &lt; -0.5 \Rightarrow 0.5 &lt; |x + 1| &lt; 1.5 \Rightarrow \frac{2}{3} &lt; \frac{1}{|x + 1|} &lt; 2.

Ok, so we have:
\left| \frac{1}{x + 1} + 1 \right| = \textcolor{blue}{\left| \frac{1}{x + 1} \right|} \textcolor{red}{|x + 2|} &lt; \textcolor{blue}{2} \textcolor{red}{|x + 2|}, ok, so we choose:

\delta = \mbox{min} \left( 0.5 , \frac{\epsilon}{2} \right).

I think your textbook's answer is, somehow, mistyped.

Ok, let's think about it, in the green part above (I mean, the restriction part), why didn't I choose: 0 &lt; |x + 2| &lt; \delta \leq 1? I didn't choose 1, that's not appropriate in this case. Can you see why?

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So, now, hint for second problem. :)

We must rearrange the expression to give k |x - a|, i.e, we must have |x - a| in our expression, so, multiply by its conjugate, we obtain.

|\sqrt{x} - 2| = \frac{\textcolor{red}{|x - 2|}}{\textcolor{blue}{| \sqrt{x} + 2 |}}

Can you go from here? :)

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EDIT: I have corrected the last LaTeX code for like 3 times, but it still shows the wrong image. >"<

Ok, now, LaTeX seems to work normally again. ^.^

 

[pardesi]

Well, nope, x cannot be a, you should add '0 <' part in front of it, like this:

well it is not necessary that x be b but there is no problem if it hiolds true for b infact then it becomes continious at that point