How Do You Know When a Algebraically System Has Infinite Solution

CuriousBanker

I intuitively and graphically understand when a function has no limit...basically if the value it approaches from opposite directions don't concur. So if from the left side it approaches negative infinity and the right side information technology approaches positive infinity, does not be...I get that. Just algebraically, can you give me a couple of examples of functions with no limits, and how I would know that without looking at a graph?

micromass

I intuitively and graphically sympathise when a part has no limit...basically if the value it approaches from reverse directions don't agree. And then if from the left side it approaches negative infinity and the right side it approaches positive infinity, does non exist...I get that.

That's not the whole story. Information technology is true that if the left-manus side and the correct-hand side limit don't agree, and then he limit doesn't be. This is called a bound discontinuity. But it can also happen that the left-manus side and right-hand side limit don't exist. In that case, it also has no limit. A famous example is ##f(x) = \sin(i/x)## in ##0##:

This is called an essential discontinuity. Those usually wait very ugly like the above.

Only algebraically, can y'all give me a couple of examples of functions with no limits, and how I would know that without looking at a graph?

It's more often than not experience. Only mayhap you can benefit from these examples:

Example 1:
Take the part
[tex]f(x) = \left\{\brainstorm{array}{l} 10~\text{if}~ten<0\\ x+2~\text{if}~x\geq 0\end{array}\right.[/tex]

This function is piecewise defined. That is, we requite a separate definition of the part in the case ##x<0## and ##x\geq 0##. In this instance, you can suspect that the limit in ##0## might not exist. Of course, some piecewise defined functions are continuous, and then it'south not because something is piecewise divers that the limitss don't be. Yous still need to check it formally by finding the left-hand side and right-hand side limits.

Some other example is

[tex]f(x) = \left\{\begin{assortment}{50} \frac{\sin(x)}{x}~\text{if}~ten\neq 0\\ 1000~\text{if}~x=0\end{array}\right.[/tex]

This function is piecewise defined. So it makes sense to doubtable the limit in ##0## doesn't exist. Even so, y'all can easily check that it in fact does exist.

Case 2:
When dealing with infinities, limits might not exist. For case, the function ##f(ten)=1/x## has no limit in ##0## since the left-hand side and correct-hand side limits don't equal.

Example 3:
When you get infinities inside a part, limits might not exist. For instance, the limit of ##f(x) = \sin(ane/ten)## doesn't exist in ##0##. A style to see this is to plug in ##0## to become ##\sin(\infty)##. This makes no sense. So this is an indication that the limit might not exist. You lot still need to cheque it rigorous.

All the same, take ##f(x) = x\sin(1/x)##. This limit does exist.

CuriousBanker

f(x)=⎧⎩⎨sin(x)x if x≠01000 if x=0 How does the limit exist in this instance? Doesn't seem like information technology does to me.

Also, why doesnt the limit of f(ten)=sin(1/x) in 0 exist? I don't fully get that one.

The other ones I get, thank you. I merely didnt know if there was a formal way of proving a limit doesnt exist or if y'all only practice it past looking at it

HallsofIvy

f(x)=⎧⎩⎨sin(x)ten if x≠01000 if ten=0 How does the limit exist in this example? Doesn't seem like it does to me.

You seem to have likewise many left braces in that! Practice y'all mean f(10)= sin(x) times x or sin(ten)/10 if x is not 0? In the outset case, the limit is 0, in the 2nd, 1.

It is important that y'all understand that in [itex]\lim_{x\to a} f(x)[/itex], the value of f(a) is completely irrelevant! The limit is the value of the part when x is arbitrarily close to a but Non equal to a. The precise argument of the definition of "[itex]\lim_{x\to a} f(x)[/itex]" is:
"[itex]\lim_{ten\to a} f(x)= L[/itex]" if and just if, for any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex], so [itex]|f(x)- L|< \epsilon[/itex]."

Look specifically at that "0< |x- a|". Nosotros are Not saying anything about what happens when |x- a|= 0, that is, when 10= a.

Besides, why doesnt the limit of f(x)=sin(one/10) in 0 exist? I don't fully get that one.

If 0< x< 0.0001, then [itex]10000< ane/x< \infty[/itex] and at that place are many intervals of length [itex]2\pi[/itex] in which sin(1/x) takes on all values from -1 to 1. The same thing happens for [itex]0< x< \delta[/itex] for whatever positive [itex]\delta[/itex]: [itex]1/\delta< x< \infty[/itex] then sin(1/x) goes (infinitely many times) from -i to 1.

The other ones I become, cheers. I just didnt know if there was a formal way of proving a limit doesnt exist or if you but do information technology by looking at it

brmath

types of functions with no limits

Looking at a graph tin give you a very proficient idea of why a limit might not exist, only in itself is never a proof. Suppose you are looking at lim f(10) as x→ a (a is a finite number). I way to prove there is no limit is to show that the right or left limit is ∞ ( or -∞). But you accept to show it -- you lot can't simply say "the graph looks that fashion". The graph may be misleading.

The example of sin(1/x) as x → 0 is one example of a grade of functions that oscillate between detached values in the vicinity of a (in this example a is 0). Yous would prove at that place is no limit in these cases by showing that the limit of one subsequence budgeted a is different from the limit of some other subsequence approaching a. In the example of sin 1/10 the outset sequence tin can be the points where sin1/ ten = i; and the second the points where sin ane/ten = 0. Both subsequences converge to 0; both have finite limits; and the limits are non the aforementioned.

Another example of the in a higher place is a part with infinitely many bespeak discontinuities every bit 10 → a. For case, you lot could have f(1/n) = i (n an integer) and f(x) = x otherwise. This role will non have a limit at 0, although it will everywhere else.

And so you meet there are several cases where a limit might not exist. Further f(x) might be a very complicated function that is hard to graph.

If yous are dealing with limits considering you lot need to get some math or physics done (as opposed to needing to pass a course), it is worth developing your skills in detecting and proving limits.

CuriousBanker

And so if I were to say "prove that the there is no limit for 1/x as ten->0, how would you prove it?

Besides, if you were to see a function like one/x equally 10->0, how would you "know" that the function tends towards infinity, and not that it doesnt just stop at some large number, like 22109123902102939230293923092913? I mean intuitively is obviosu but how do you prove it?

brmath

Well, I would say, consider the sequence 1/due north. If f(10) = 1/x then f(1/due north) = northward. Since yous can pick n as large as you similar y'all can satisfy the strict mathematical definition of the limit being infinity. That says, basically that if you cull any stock-still number North, you can e'er detect an 10 such that f(x) >North. The essential point is that the North has to exist picked offset. Then you selection the N, I'll pick x = 1/(N+ane), and f(ten) > N.

Now you do have to worry well-nigh x's that are between one/due north and ane/(n+1). Theoretically it could be that there is a sequence of a's going to 0 where f(a) = i for every a, or something of that kind. If it could happen, so we actually do non have a limit. Just in the example of 1/10 that cannot happen, considering 1/ten is continuous. We cannot fifty-fifty jiggle 1/x into having peaks and valleys on its way to infinity, considering 1/x is monotonic.

How Do You Know When a Algebraically System Has Infinite Solution

How to detect algebraically when a function has no limit?

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