Question 988009
<pre>
We are to prove that no matter how small a tolerance <font face="symbol">e</font> > 0 we desire for
(5-2x) to be within -1, we will always be able to find at least one value 
for <font face="symbol">d</font> > 0 such that whenever x is within <font face="symbol">d</font> of 3, that (5-2x) will always be 
within <font face="symbol">e</font> of -1.  

Suppose we are given <font face="symbol">e</font> > 0 and desire to prove that there exists a value of 
<font face="symbol">d</font> > 0 such that |f(x)-(-1)| < <font face="symbol">e</font>.  We look at the absolute value of the 
difference between the (5-2x) and -1

|(5-2x)-(-1)| = |5-2x+1| = |6-2x| = 2|3-x| = 2|x-3| we want to be < <font face="symbol">e</font>

So the (5-2x) will be within <font face="symbol">e</font> of -3 if 2|x-3| < <font face="symbol">e</font>, which is to say

when |x-3| < <font face="symbol">e</font>/2, so if we choose any <font face="symbol">d</font> < <font face="symbol">e</font>/2, then when |x-3| < <font face="symbol">d</font>, that is,
when x is within <font face="symbol">d</font> of 3, (5-2x) is within <font face="symbol">e</font> of 3.                                    

So given <font face="symbol">e</font> > 0 |(5-2x)-(-1)| < <font face="symbol">e</font> whenever <font face="symbol">d</font> < <font face="symbol">e</font>/2.

That proves that 

{{{matrix(2,4,

lim,(5-2x),""="",-1,
"x->3","","","")}}}

Edwin</pre>