When we try to find a one-sided limit, we use a less mathematical\algebraic approach? Either graph it, or maybe try to plug some close numbers from a corresponding side?
Aha, NOW I understand your first question.
In terms of "finding" a PROVISIONAL answer, any method will do: graphing, analogy, trial and error, intuition, prayer, or sorcery. Some methods are more efficient than others of course. But until you prove by logic that your provisional answer is correct, you have not "found" the correct answer. The difference between the two meanings of "find" is what Popper called the difference between the psychology of discovery and the logic of scientific discovery.
You can get an excellent provisional answer in this case by graphing, but a graph is a not a proof.
Moving on, your reference to derivatives indicates that you are trying to use L'Hospital's Rule. That rule does not apply to this case. (To see why, try deriving the rule; it applies only in two special cases.)
Although this is seldom pointed out explicitly, the laws of limits also apply to one-sided limits.
$\displaystyle \lim_{x \rightarrow c}f(x) \in \mathbb R,\ \lim_{x \rightarrow c}f(x) \ne 0,\ and\ \lim_{x \rightarrow c}g(x) = 0 \implies \lim_{x \rightarrow c}\dfrac{f(x)}{g(x)} = what?$$
Why is there any problem at all? taking x= 0 in the fraction, we get . That tells us immediately that the fraction will grow without bound as x goes to 0. (In fact in this problem, the "going to 0 from above" is irrelevant- we would get the same answer for x going to 0 from below or just "x going to 0".)
There is no limit. (or "the limit is " which just says there is no limit in a specific way.)
(More generally, if setting x= 0 (or whatever x is tending to) in a fraction, we get with b non-zero, then that is the limit. If b= 0 and a is not, there is no limit. It is only when a= b= 0 that we need to look closer.)
Okay, so showing that the denominator is equal to zero is enough to show that the limit is some infinity?
And I mean this all includes some verbal justification... In the case of some number not equal to zero divided by zero limits from two different sides go to different infinities. We can show it by plugging values close to zero from either side. That is what my question was basically about (it's about one-sided limits, but the gist of the question is basically the same). It seems like this reasoning is implied somehow and not really put in mathematical terms when we solve these limits.
And also, it seems like the limit to 0 from the negative side would be minus infinity in the above example? So how is this shown in a strict mathematical way? Or it shouldn't be?
http://www.wolframalpha.com/input/?i...2+-1%29+x+to+0
Interestingly, Wolfram doesn't show steps for one-sided limits
Oh, right! I didn't look closely enough. If x> 1 then x+ 1> 1 so [tex]\sqrt{x+ 1}> 1 [tex], and both numerator and denominator are positive. But if x< 1, x- 1< 1 so , . now the numerator is still positive but the denominator is negative so the fraction is negative. I would still say that the limit does not exist.
Some texts will say:
$\displaystyle \lim_{x\to 0^+} \dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} - 1} = \infty$
This means something a bit special.
First of all, $\infty$ is not, nor will it ever be, a real number. The above is an ABBREVIATION for a rather technical condition.
Informally, we mean as $x$ "approaches" (gets close to) 0 from the right, the fraction we are trying to find the limit of, "grows without bound". So, in a very "real" (excuse the pun :P) sense, there is no limit.
Nevertheless, we can say precisely what we mean, without even MENTIONING infinity. Technically, this is what we mean:
For any $M \in \Bbb N^+$, there IS (exists) a real number $\delta > 0$ such that:
$0 < x < \delta$ forces $\dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} - 1} > M$ <---we don't need the usual absolute value signs in the "delta part" since we are approaching 0 "from the positive side", and we don't need the absolute values signs for the "M part", since M is positive.
So to PROVE the above statement, given $M$, we have to find the $\delta$.
Let's do some algebraic wizardry, first:
$\dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} - 1} = \dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} - 1}\cdot \dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}$
$= \dfrac{(\sqrt{x+1} + 1)^2}{x+1 - 1} = \dfrac{(x+1) + 2\sqrt{x+1} + 1}{x} = \dfrac{x + 2\sqrt{x+1} + 2}{x}$
$= 1 + \dfrac{2\sqrt{x+1}}{x} + \dfrac{2}{x} > \dfrac{2}{x}$, since each term in the sum is positive, when $x > 0$.
Now suppose we are given $M$ (this might be a very large $M$).
Choose $\delta = \dfrac{2}{M}$.
If $0 < x < \dfrac{2}{M}$, then:
$0 < \dfrac{M}{2} < \dfrac{1}{x}$, so that:
$M = \dfrac{2M}{2} < \dfrac{2}{x} < 1 + \dfrac{2\sqrt{x+1}}{x} + \dfrac{2}{x} = \dfrac{\sqrt{x+1} + 1}{\sqrt{x+1} - 1}$, QED.
They are equivalent conditions, although this is not immediately apparent. If $f(x) > M$ for any positive integer $M$, then $f(x) > a$ for any $a \in (0,M)$, so that $f(x) > y$ for any positive real number $y$ that is:
a) either an integer,or
b) in-between two integers
which pretty much covers the possibilities.
On the other hand, if $f(x) > M$ for any real number $M$, then it is certainly greater than $\lfloor M \rfloor$, which is an integer. As $M$ progresses through every real positive number, $\lfloor M \rfloor$ goes through every positive integer.
(This is a consequence of the "Archimedean property" of the real numbers, and it is VERY USEFUL).
As to why we singled out 2/x:
If we want to show that something gets big, bigger than any integer can get, one way to show this happens is to show it is bigger than something else we KNOW ALREADY gets big.
It helps if the thing (expression, function, etc.) we are bigger than has a nice simple form that is easy to work with. Since the square root stuff is difficult to calculate with, we want something easier to deal with.
Once I decided that our function (the fraction with the square roots top and bottom) was bigger than 2/x, the next thing was to find a way to choose δ so that 2/δ ≥ M.
That gave me the expression for δ I needed: δ = 2/M. Then I started writing the proof, sort of in "reverse order" from how I solved it, if that makes any sense.
So what are the implication of this in our case? Or did you just want to simply point it out?
And as to why I was confused about N^+, is the plus looked redundant (well, apart from the fact that I also thought it was supposed to be R^+ in any case, but you already cleared this up). As far as I understand Z^+ = N
I just wanted to point out that if you see an unbounded argument based on an $M$ that is a real number, you can convert it into one where $M$ is an integer (and also vice-versa).
As for the plus sign, some people include 0 in the natural numbers; I just wanted to make it perfectly clear our integer was positive, not merely non-negative. I could have used $\Bbb Z^+$, no big deal.