Question 1044993
.
Example: Please help me solve this equation:  if f(x)={{{ (3x^2+x)/2x }}} for x=!0
                                                  {{{f(0)=k}}}
and if ƒ is continuous at x = 0, then k =
A) -3/2
(B) -1
(C) 0
(D) 1
(E) 3/2 answer here
~~~~~~~~~~~~~~~~~~~~~


<pre>
The formulation is incorrect. (And the solution is incorrect as well).

The correct formulation and solution follow:


    You are given a function  f(x) = {{{(3x^2+x)/(2x)}}}.
    Find the limit of the function at x-->0.


<U>Solution</U>. 


Factor x out parentheses in the numerator. Then cancel factor "x" in the numerator an denominator. You will get

lim   {{{(3x^2+x)/(2x)}}} =  lim  {{{(x*(3x+1))/(2x)}}} =  lim   {{{(3x+1)/2}}} = {{{1/2}}}.
x-->0              x-->0               x-->0

Notice that this result  lim = {{{1/2}}}  is not in the list.
Nevertheless, this result is correct.
Everything in the list is incorrect.
</pre>

The plot illustrates it:


<TABLE> 
  <TR>
  <TD> 

{{{graph( 330, 330, -3.5, 3.5, -3.5, 3.5,
          (3x^2+x)/(2x)
)}}}


Plot y = {{{(3x^2+x)/(2x)}}}

  </TD>
  </TR>
</TABLE>