sry mistyped that question

See reply #3 to your original post.

x tends to 0 ((3+2x)^5-243)/(3x)

$\begin{array}{|rcll|} \hline && \lim \limits_{x\to 0} { \left( \frac{ (3+2x)^5-243 } {3x} \right) } \\\\ &=& \lim \limits_{x\to 0} { \left( \frac{ \binom{5}{0}3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -243 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ 3^5+ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 -3^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( \frac{ \binom{5}{1}3^4(2x)+ \binom{5}{2}3^3(2x)^2+ \binom{5}{3}3^2(2x)^3+ \binom{5}{4}3(2x)^4+ \binom{5}{5}(2x)^5 } {3x} \right) }\\\\ &=&\lim \limits_{x\to 0} { \left( ~ \binom{5}{1}3^3\cdot 2+ \binom{5}{2}3^2\cdot 4x+ \binom{5}{3}3^1\cdot 8x^2+ \binom{5}{4}16x^3+ \binom{5}{5}\frac{32}{3}x^4 ~ \right) }\\\\ &=& \binom{5}{1}3^3\cdot 2 \\\\ &=& 5\cdot 27 \cdot 2 \\\\ &=& 270 \\\\ \hline \end{array}$