Question 995368
<pre>
They don't know what you want.  Here's the solution.

{{{matrix(2,2,

lim,(2^t-8)/(t-3),
"t->3","")}}}

Don't just think derivative, think L'Hopital's rule.  Since both
numerator and denominator approach 0, L'Hopital's rule applies:

To find the derivative of 2<sup>t</sup>, we use this formula: {{{d(a^u)/dx=a^u*ln(u)*"u'"}}}

{{{matrix(2,2,

lim,(2^t*ln(2))/(1),
"x->3","")}}}

{{{matrix(2,2,

lim,2^t*ln(2),
"x->3","")}}}

{{{2^3*ln(2)}}}

{{{8ln(2)}}}

Edwin</pre>