<PRE><B>Question 1118241</B>
&lt;br&gt;
A couple of preliminary comments....&lt;br&gt;
(1) Do you really think it is any help to us to include &quot;HINT [See Example 2.]&quot; in your message to us?!&lt;br&gt;
(2) The problem as you show it is of little interest, because it is impossible to define f(a) to make f continuous at x=0:&lt;br&gt;
{{{f(0) = 2-2e^0/0 = 2-1/0}}} which is undefined.&lt;br&gt;
If you are working on a problem like this, then your knowledge of mathematics should be sufficient that you would know appropriate use of parentheses is critical.&lt;br&gt;
The function you are asking about is, in fact, {{{f(x) = (2-2e^x)/x}}}; not {{{f(x) = 2-2e^x/x}}}.&lt;br&gt;
At x=0, the function value is the indeterminate form 0/0.&lt;br&gt;
So use l&#39;Hospital&#39;s rule.&lt;br&gt;
limit as x approaches 0 of {{{(2-2e^x)/x}}}&lt;br&gt;
is equal to&lt;br&gt;
limit as x approaches 0 of {{{-2e^x/1}}} = -2e^0 = -2&lt;br&gt;
A graphing calculator will confirm this answer.&lt;br&gt;
{{{graph(400,400,-5,5,-10,2,(2-2e^x)/x)}}}</PRE>