Finding Limits

**^_^Engineer_Adam^_^** · December 19th 2006, 03:26 PM

This came on my pre midterm yesterday
and this is the only problem i cant answer:

Find the Limit
$\lim_{x\to 0} \frac{x+1}{|x|}$

Thank you very much helpers!
Happy holidays!!

**galactus** · December 19th 2006, 03:46 PM

Rewrite as:

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

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

Apparently, Latex is down again. What does "can not open stream", no such file or directory" mean?.

**AfterShock** · December 19th 2006, 03:47 PM

Originally Posted by ^_^Engineer_Adam^_^

This came on my pre midterm yesterday
and this is the only problem i cant answer:

Find the Limit
$\lim_{x\to 0} \frac{x+1}{|x|}$

Thank you very much helpers!
Happy holidays!!

Infinity. Try graph it and see what happens.

**ThePerfectHacker** · December 19th 2006, 05:17 PM

Originally Posted by ^_^Engineer_Adam^_^

This came on my pre midterm yesterday
and this is the only problem i cant answer:

Find the Limit
$\lim_{x\to 0} \frac{x+1}{|x|}$

Thank you very much helpers!
Happy holidays!!

Consider the right handed limit,
$\lim_{x\to 0^+}\frac{x+1}{|x|}$
In this case,
$x>0$ thus, $|x|=x$ .
Thus,
$\lim_{x\to 0^+}1+\frac{1}{x}$
The rest is trivial.
This limit does not exist.
Thus, the orginal limit
$\lim_{x\to 0}\frac{x+1}{|x|}$
Does not exist.
Because we require for the left-right handed limits to exist.

**AfterShock** · December 19th 2006, 06:53 PM

Originally Posted by ThePerfectHacker

Thus, the orginal limit
$\lim_{x\to 0}\frac{x+1}{|x|}$
Does not exist.

I disagree. Although the lim(x->0) 1/x is undefined, 1/abs(x) is not.

I still think it's infinity.

**ThePerfectHacker** · December 19th 2006, 07:39 PM

Originally Posted by AfterShock

I disagree. Although the lim(x->0) 1/x is undefined, 1/abs(x) is not.

I still think it's infinity.

The limit does not exist.
You cannot choose a $\delta$ for each $\epsilon$ , surly.

**TD!** · December 20th 2006, 01:06 AM

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):

$ \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.

**galactus** · December 20th 2006, 03:15 AM

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:

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

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

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

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

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'?.

**ThePerfectHacker** · December 20th 2006, 07:01 AM

Originally Posted by TD!

$ \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 $

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).

**TD!** · December 20th 2006, 07:05 AM

Originally Posted by ThePerfectHacker

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).

Perhaps you didn't mean to say it like this, but in the definition I gave, the limit does exist and is +infinity (by definition!).

**galactus** · December 20th 2006, 07:25 AM

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.

**AfterShock** · December 20th 2006, 09:11 AM

Originally Posted by galactus

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.

Huh? 1/x^2? This is clearly infinity as x approaches 0. I'll agree with the other two.

**TD!** · December 20th 2006, 09:41 AM

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):

$ \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:

$ \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"

**galactus** · December 20th 2006, 10:19 AM

Originally Posted by AfterShock

Huh? 1/x^2? This is clearly infinity as x approaches 0. I'll agree with the other two.

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.

**TD!** · December 20th 2006, 10:22 AM

That's just a matter of definition, see above

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