If $f(x) \le g(x)$ for all $x$ on $[a,b]$ (except possibly at $x = c$) and $a \le c \le b$ then $$ \lim_{x \to c} f(x) \le \lim_{x \to c} g(x)$$

May I know what this means?

Question

If $f(x) \le g(x)$ for all $x$ on $[a,b]$ (except possibly at $x  = c$) and $a \le c \le b$ then $$ \lim_{x \to c} f(x) \le \lim_{x \to c} g(x)$$ 

May I know what this means?

parthkohli · Answer

Is this as simple as something like 'if $f(x)$ is less than or equal to $g(x)$, then we have $\lim \limits_{x \to c} f(x) \le \lim \limits_{x \to c} g(x)$'?

anonymous · Answer

yes

anonymous · Answer

you want a proof for this or what ?!

parthkohli · Answer

No, I don't. I just saw this on Paul's Notes and I got a little confused as to what it actually was supposed to mean.

anonymous · Answer

not much to this, don't think too hard about it

parthkohli · Answer

What's that $[a,b]$ and $c$ thing? Are [a,b] the coordinates?

anonymous · Answer

[a,b] is the set of the real numbers between a and b including a and b !

parthkohli · Answer

Oh! That's the interval notation!

parthkohli · Answer

Wait. How can we have $x \in [a,b], \quad  x \ne c, \quad  a \le  b \le c$ at the same time?

parthkohli · Answer

Since $x$ fulfils the value of numbers between $a$ and $b$; and $c$ is between a and b, then $c$ also is a solution!

parthkohli · Answer

Oh. I get it now :)

parthkohli · Answer

$x \in [a, c) \cup (c,b]$.

anonymous · Answer

yeaah !