# real analysis one sided limits

• Oct 2nd 2012, 02:10 PM
biga415
real analysis one sided limits
hey guys. so im in an introductory real analysis course and I am finding it quite challenging, partially because my teacher isnt great. i went to ask for help on some limit problems but im not sure if his advice was accurate.

1) Find lim(x approaches x0-) f(x) and lim((x approaches x0+) f(x) for f(x)=(x+abs(x))/x x0=0 i found the solutions (2, x>0 and 0, x<0) easily. The book asks for an epsilon-delta proof if possible and my teacher claims it is trivial because they are constant. if this is the case when should you finish the proof?

2) I would also appreciate help with a second problem that asks the same thing except when f(x)= abs(x-1)/(x2 +x-2), x0=1

3) prove: if limf(x) exists, there is a constant M and a p>0 such that abs(f(x)) < or = M if 0<abs(x-x0)<p. I was having trouble with this one also. in this case, basically p is delta right? and if we assume f(x)=M, then abs(M-L)=epsilon. should i approach this proof with a contradiction?

• Oct 2nd 2012, 02:48 PM
Plato
Re: real analysis one sided limits
Quote:

Originally Posted by biga415
1) Find lim(x approaches x0-) f(x) and lim((x approaches x0+) f(x) for f(x)=(x+abs(x))/x x0=0 i found the solutions (2, x>0 and 0, x<0) easily. The book asks for an epsilon-delta proof if possible and my teacher claims it is trivial because they are constant. if this is the case when should you finish the proof?

2) I would also appreciate help with a second problem that asks the same thing except when f(x)= abs(x-1)/(x2 +x-2), x0=1

3) prove: if limf(x) exists, there is a constant M and a p>0 such that abs(f(x)) < or = M if 0<abs(x-x0)<p. I was having trouble with this one also. in this case, basically p is delta right? and if we assume f(x)=M, then abs(M-L)=epsilon. should i approach this proof with a contradiction?

I for one find this almost impossible to read!
It looks like the first question is:
$\lim _{x \to 0^ - } \frac{{x + |x|}}{x}\;\& \,\lim _{x \to 0^ + } \frac{{x + |x|}}{x}$

If you click on the "reply with quote" tab you will have access to the LaTeX code.
• Oct 2nd 2012, 07:27 PM
biga415
Re: real analysis one sided limits
haha, sorry, i dont know how to use LaTex, however, you did get the problem correct!
• Oct 2nd 2012, 07:37 PM
biga415
Re: real analysis one sided limits
#2
$f(x)=\frac{{|x-1|}}{}{x^{2}+x-2}$
• Oct 3rd 2012, 03:19 AM
Plato
Re: real analysis one sided limits
Quote:

Originally Posted by biga415
dont know how to use LaTex, however, you did get the problem correct!

$\frac{{x + \left| x \right|}}{x} = \left\{ {\begin{array}{rr} {0,} & {x < 0} \\ {2,} & {x > 0} \\\end{array}} \right.$
This makes #1 easy.
• Oct 3rd 2012, 04:46 AM
johnsomeone
Re: real analysis one sided limits
For #2, factor the denominator, then try to do what Plato did in the 1st example. When x>1, what's f(x)? When x<1, what's f(x)? Once you understand that, determining the two one-sided limits will hopefully be easy.
• Oct 3rd 2012, 05:40 AM
biga415
Re: real analysis one sided limits
One of my main questions was whether or not I need to do the epsilon-delta proof at all for number one. my professor claims i dont, but if thats the case when should I? also would i show two one-sided limit proofs?
• Oct 3rd 2012, 05:59 AM
johnsomeone
Re: real analysis one sided limits
Yes, I think that's what he's asking for, an epsilon-delta proof for each of the one-sided limits.

If you recall, the full limit speaks of $\forall x s.t. 0 < | x - a | < \delta$, which is $\forall x \in (a-\delta, a) \cup (a, a+\delta)$.

The one sided limits are defined exactly the same, except applying to only one of those halves:

For $\lim_{x \to a^-}$ it's $\forall x \ s.t. \ a - \delta < x < a$, which is $\forall x \in (a-\delta, a)$.

For $\lim_{x \to a^+}$ it's $\forall x \ s.t. \ a < x< a + \delta$, which is $\forall x \in (a, a+\delta)$.

In full:

$\lim_{x \to a^-}f(x) = L$ iff $\forall \epsilon > 0 \ \exists \delta > 0 \ni a - \delta < x < a \Rightarrow \lvert f(x) - L \rvert < \epsilon$.

$\lim_{x \to a^+}f(x) = L$ iff $\forall \epsilon > 0 \ \exists \delta > 0 \ni a < x < a+ \delta \Rightarrow \lvert f(x) - L \rvert < \epsilon$.