I Limit of Extension: Can Function Have Different Limit?

[Danijel]

When we define a limit of a function at point c, we talk about an open interval. The question is, can it occur that function has a limit on a certain interval, but it's extension does not? To me it seems obvious that an extension will have the same limit at c, since there is already infinitely many points around its limit L, so adding more, or none, doesn't change anything (since we are now looking at sequences whose limit is c, but end up in the extension's domain,which is larger than the original domain, so we have more sequence values, of course, if we don't talk about Cauchy definition). However, my short experience with calculus has made me careful of concluding wacky statements (at least for those Iack I a firm argument).

Edit: I note that a function can behave differently on a different interval, hence it can happen that, for some small number δ, there is such c_n for which f(c_n) exits the interval <L-δ,L+δ>, and the function (the extension) does not have a limit L at c.
But, since for large n, by our assumption that the sequence c_n has a limit at c, for a large n, the f(c_n) cannot exit the observed interval, since preimage is already in a small interval around c, and for every such point, it holds that |f(c_n)-L|<δ. So a function may have a limit. I really can't think straight.

 

[FactChecker]

It's not clear to me what type of function, domain, range, and extension you are talking about.
I will assume that you are talking about the analytic extension of a real-valued function of a real variable to complex analytic function on the complex plane. In that case, the analytic function can have a wide variety of behaviors off the real line. Although the limit of the real function would be at least a cluster point of the complex function, it may not be the limit of the complex function.

 

[Danijel]

I gave little information, and I am sorry.
To skip the settings, here's straight to the problem. Say we want to prove that limit of the function f(x)=sinx/x as x approaches 0 is 1. We can play around and get that cosx<sinx/x<1 for 0<x<π/4. Since the limit of cosx as x approaches 0 is 1, and same goes for the constant function h(x)=1, by the squeeze theorem, we have our proof. However, the squeeze theorem talks only about the function (at least as I am aware of) whose domain is the interval for which cosx<sinx/x<1 holds, and that is (because of the evenness of functions cosx and x) <-π/4,0>U<0,π/4>. So the function has a limit in this interval. But f(x)=sinx/x is defined on ℝ\{0}. What I am asking is, if we extend <-π/4,0>U<0,π/4> to the function's natural domain, does the function has the same limit.

 

[FactChecker]

Suppose you extend the function to include x=0:
\begin{equation}
f(x)=\begin{cases}
sin(x)/x, & \text{if $x \neq 0$}.\\
1, & \text{if $x = 0$}.
\end{cases}
\end{equation}

Then the extended function is defined, continuous, and has a limit of 1 at x=0 by the usual definitions.

 

[mfb]

If you want to consider the limit of f(x)=sinx/x in the real numbers, you have to consider both sides. Otherwise you get a single-sided limit.

Consider this function:$$\begin{equation}
g(x)=\begin{cases}
sin(x)/x, & \text{if $x > 0$}.\\
1, & \text{if $x = 0$}. \\
0, & \text{if $x<0$}.
\end{cases}
\end{equation}$$It has the same behavior for positive x, but overall its limit for x->0 does not exist.

The same can happen if you go to complex numbers, and here the examples are easier to write down: $\displaystyle h(x)=e^{-1/x^2}$
Continuous everywhere in the real numbers, and converges to 0 for x->0, but in the complex numbers this function goes crazy there.

 

[mathman]

I gave little information, and I am sorry.
To skip the settings, here's straight to the problem. Say we want to prove that limit of the function f(x)=sinx/x as x approaches 0 is 1. We can play around and get that cosx<sinx/x<1 for 0<x<π/4. Since the limit of cosx as x approaches 0 is 1, and same goes for the constant function h(x)=1, by the squeeze theorem, we have our proof. However, the squeeze theorem talks only about the function (at least as I am aware of) whose domain is the interval for which cosx<sinx/x<1 holds, and that is (because of the evenness of functions cosx and x) <-π/4,0>U<0,π/4>. So the function has a limit in this interval. But f(x)=sinx/x is defined on ℝ\{0}. What I am asking is, if we extend <-π/4,0>U<0,π/4> to the function's natural domain, does the function has the same limit.

To get a limit at 0, we need to consider what happens only in a small interval around 0. The extension you are considering is irrelevant.

 

[Stephen Tashi]

However, the squeeze theorem

Terminology such as "the squeeze theorem" is slightly ambiguous because different texts can state "the squeeze theorem" different ways. How do your text materials state the theorem?

talks only about the function (at least as I am aware of) whose domain is the interval for which cosx<sinx/x<1 holds,

A typical statement of the squeeze theorem is that $g(x) \le f(x) \le h(x)$ holds for each $x$ in an open interval $(a - \delta, a + \delta)$ except perhaps at $x = a$. So its conclusion about $lim_{x \rightarrow a} f(x)$ does not require that the inquality holds at $x = a$.

However, my short experience with calculus has made me careful of concluding wacky statements (at least for those Iack I a firm argument).

That's the correct outlook for both wacky statements and intuitively plausible statements in mathematics. However, it puts you in a state of perpetual uncertainty unless you also develop confidence about judging whether general theorems apply to specific situations. To apply general theorems, you have develop a legalistic skill in paying attention to what the hypotheses of theorems are.

Ideally, a person should have a skeptical outlook about a general theorem and relieve it by reading the proof. Then the theorem is applied confidently afterwards. Of course, nobody has such ideal behavior.