I have this question here:

"Suppose that a function f(x) is defined for all x in [-1,1]. Can anything be said about the existence of lim[x->0] f(x)? Give reasons for your answer."

I don't think I understand the description too well. For [-1,1] I imagine a horizontal line running from -1 to 1 at y = 0. If that's the case then as x approaches 0 the limit would be 0 right? This is my first semester of calculus and this stuff is really confusing to me. Am I approaching this problem correctly?

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    $\begingroup$ FIne, but you have only considered one simple function. What about a function that is $-1$ for all $x < 0$ and $+1$ otherwise? $\endgroup$ – Mark Fischler Jan 19 '16 at 5:28
  • $\begingroup$ That is a good point. I remember having a problem like that in class; the limit wouldnt exist if that was the case. $\endgroup$ – Xirol Jan 19 '16 at 5:31
  • $\begingroup$ However, my problem implies that it's one function and x is everything within [-1,1]. So would my answer be more fitting? $\endgroup$ – Xirol Jan 19 '16 at 5:34

It seems to me that you are a bit unclear on the concept of function, as well as what the question is asking.

The statement "$f(x)$ is a function defined for all $x$ in $[-1,1]$" can be thought of in this way: $f(x)$ is a rule which makes sense, and gives one definite answer, for any value of $x$ from $-1$ to $1$. For example:

  • $f(x)=x^2$ defines a function, because for each specific $x$ in $[-1,1]$ you can calculate $x^2$, and you will get one definite answer;
  • $f(x)=2x-7$ defines a function for essentially the same reason;
  • $f(x)=1/x$ does not define a function on $[-1,1]$, because you cannot calculate a value for $f(0)$;
  • $f(x)=1/(x+2)$ defines a function on $[-1,1]$, because if $x$ is between $-1$ and $1$ then $x+2$ cannot be $0$ and there is no problem dividing;
  • $f(x)=\pm\sqrt{x+2}$ does not define a function on $[-1,1]$, because for any $x$ value you will get a choice of two answers, not one definite answer;
  • $f(x)=\cases{1&if $-1\le x<0$\cr 2&if $0\le x\le1$\cr}$ defines a function since for any given $x$ you can calculate one definite value of $f(x)$.

Note: in more advanced courses you will get a more detailed and precise concept of what is meant by a function. But the above will do for now.

The question asks: for functions such as those we have described, will $\lim_{x\to0}f(x)$ exist? - yes? or no? or sometimes yes and sometimes no?

The example you have given is $f(x)=0$. (More or less: you didn't actually say it was the function $f$, you said it was the interval $[-1,1]$, which is one of the things that makes me think you may not understand the function concept perfectly.)

But say it's $f(x)=0$: then as you have pointed out, the limit exists and is equal to $0$. So this rules out the "no" answer, and you have to decide between "yes" and "sometimes". So, can you find a function where the limit does not exist? Hint: look at the examples I have given above.

Good luck!

  • $\begingroup$ You were correct that I dont quite understand functions fully. Thank you so much for helping me understand. I really appreciate it David! :) $\endgroup$ – Xirol Jan 19 '16 at 6:08

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