This came on my pre midterm yesterday

and this is the only problem i cant answer:

Find the Limit

$\displaystyle \lim_{x\to 0} \frac{x+1}{|x|}$

Thank you very much helpers!

Happy holidays!!

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- Dec 19th 2006, 03:26 PM^_^Engineer_Adam^_^Finding Limits
This came on my pre midterm yesterday

and this is the only problem i cant answer:

Find the Limit

$\displaystyle \lim_{x\to 0} \frac{x+1}{|x|}$

Thank you very much helpers!

Happy holidays!! - Dec 19th 2006, 03:46 PMgalactus
Rewrite as:

$\displaystyle \lim_{x\to 0}(x+1)\lim_{x\to 0}\frac{1}{|x|}$

$\displaystyle \lim_{x\to 0}\frac{1}{|x|}=\infty$

Apparently, Latex is down again. What does "can not open stream", no such file or directory" mean?. - Dec 19th 2006, 03:47 PMAfterShock
- Dec 19th 2006, 05:17 PMThePerfectHacker
Consider the right handed limit,

$\displaystyle \lim_{x\to 0^+}\frac{x+1}{|x|}$

In this case,

$\displaystyle x>0$ thus, $\displaystyle |x|=x$.

Thus,

$\displaystyle \lim_{x\to 0^+}1+\frac{1}{x}$

The rest is trivial.

This limit does not exist.

Thus, the orginal limit

$\displaystyle \lim_{x\to 0}\frac{x+1}{|x|}$

Does not exist.

Because we require for the left-right handed limits to exist. - Dec 19th 2006, 06:53 PMAfterShock
- Dec 19th 2006, 07:39 PMThePerfectHacker
- Dec 20th 2006, 01:06 AMTD!
It just depends on your definition of the limit. In the 'classical' definition, this would indeed be "does not exist". You can extend that definition for improper limits, I'll give one example (applicable to this case):

$\displaystyle

\mathop {\lim }\limits_{x \to a} f\left( x \right) = + \infty \Leftrightarrow \forall \alpha \in \mathbb{R} ,\exists \delta > 0:0 < \left| {x - a} \right| < \delta \Rightarrow f\left( x \right) > \alpha

$

You can do the same for -inf and for limits where you let x approach +/- inf. - Dec 20th 2006, 03:15 AMgalactus
Thanks for the fix up, PH. I checked and checked my code. I kept getting these error messages, though.

Anyway, I had thought of just rewriting it this way:

$\displaystyle \lim_{x\rightarrow{0}}(x+1)$$\displaystyle \lim_{x\rightarrow{0}}\frac{1}{\sqrt{x^{2}}}$

$\displaystyle |x|=\sqrt{x^{2}}$

$\displaystyle \lim_{x\rightarrow{0^{-}}}\frac{1}{\sqrt{x^{2}}}={\infty}$

$\displaystyle \lim_{x\rightarrow{0^{+}}}\frac{1}{\sqrt{x^{2}}}={ \infty}$

Also, for a check,I ran it through Maple. It says infinity also.

**But then again, it appears it is unbounded as x approaches 0, therefore, the limit can not exist.**

Calculators and the like will still give infinity as the answer, though.

As TD says, it depends on your definition. Is this what could be known as a 'subjective limit'?. - Dec 20th 2006, 07:01 AMThePerfectHacker
My differencial equations professor used to say "... if the limit exists and is finite...". I should have asked him after class why he used that phrase. But I assumed he used a different definition for an existing limit. I just feel safer using the classical definition.

Now, it can be shown (I believe I shown this some time ago) that an infinite limit (the definition you gave) never exists (meaning you cannot find a delta for each epsilon). - Dec 20th 2006, 07:05 AMTD!
- Dec 20th 2006, 07:25 AMgalactus
As PH said, a definition I was always taught was "If f(x) does not approach a single number as x approaches c, then the limit does not exist?".

sin(1/x) is a classic example. Along with 1/x^2 and |x|/x. - Dec 20th 2006, 09:11 AMAfterShock
- Dec 20th 2006, 09:41 AMTD!
Perhaps this is a good time to elaborate a bit on "the definition" of a limit.

For taking the limit, we could do six different things (*a*is a real number):

$\displaystyle

\mathop {\lim }\limits_{x \to a} \, , \, \mathop {\lim }\limits_{x\mathop \to \limits^ > a} \, , \, \mathop {\lim }\limits_{x\mathop \to \limits^ < a} \, , \, \mathop {\lim }\limits_{x \to + \infty } \, , \, \mathop {\lim }\limits_{x \to - \infty } \, , \, \mathop {\lim }\limits_{x \to \infty }

$

Number 2&3 are a right and left (or upper and lower) limit respectively, sometimes written differently.

With the last statement, I mean: we let x go to infinity but we don't care about its sign, so |x| -> +infinity.

That was taking the limit, but we can also get (four) different results:

$\displaystyle

\lim f\left( x \right) = b \, , \, \lim f\left( x \right) = + \infty \, , \, \lim f\left( x \right) = - \infty \, , \, \lim f\left( x \right) = \infty

$

Here,*b*is a real number and the last one again means unbounded, so |f(x)| larger than any real number.

Combining the six ways of taking the limit and the four possible results, you actually get 24 definitions!

Of course, they only make sense in the extended reals, since we're also dealing with infinity.

I included them under here in an image, to not overload LaTeX.

You'll recognise the first one as the "classic definition" :)

http://img105.imageshack.us/img105/1928/limdefxk9.gif - Dec 20th 2006, 10:19 AMgalactus
Hello Aftershock:

After reading your post, I happened to find this particular limit is a calc book.

I have attached the scan from it. That's what it says. Just pointing it out.

Anyway, this particular text goes by the definition that if the limit does not approach a number, then the limit does not exist. Infinity is not a number, therefore, the limit does not exist. That appears to be the consensus in many places I have read from--both text and web. - Dec 20th 2006, 10:22 AMTD!
That's just a matter of definition, see above ;)