An Infinite Limit? The following question has a finite limit $\left( {{{11e} \over {24}}} \right)$, which can be easily obtained from the method of expansions. But I get an infinite limit, and I am not exactly sure where I have gone wrong. Please let me know also the reason why a certain step cannot be undertaken.
Question - 
$$Evaluate\,\mathop {\lim }\limits_{x \to 0} {{{{(1 + x)}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}} - e + {\textstyle{1 \over 2}}ex} \over {{x^2}}}$$
My Try - 
$$\eqalign{
  & Let\,\,L = \mathop {\lim }\limits_{x \to 0} {{{{(1 + x)}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}} \over {{x^2}}} - \mathop {\lim }\limits_{x \to 0} {{e - {\textstyle{1 \over 2}}ex} \over {{x^2}}}  \cr 
  & Let\,\,{L_1} = {e^{\mathop {\lim }\limits_{x \to 0} \log \left( {{{{{(1 + x)}^{{\textstyle{1 \over x}}}}} \over {{x^2}}}} \right)}}\,\,and\,{L_2} = {e^{\mathop {\lim }\limits_{x \to 0} \log \left( {{{e - {\textstyle{1 \over 2}}ex} \over {{x^2}}}} \right)}}  \cr 
  & {L_1} = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{\log (1 + x)} \over x} - \log ({x^2})} \right)}}\,\,and\,{L_2} = {e^{\mathop {\lim }\limits_{x \to 0} \log \left( {e\left( {1 - {\textstyle{x \over 2}}} \right)} \right) - \log {x^2}}}  \cr 
  & {L_1} = {e^{1 - \mathop {\lim }\limits_{x \to 0} \log {x^2}}}\,and\,\,{L_2}\, = \,{e^{1 + }}^{\mathop {\lim }\limits_{x \to 0} \log \left( {1 - {\textstyle{x \over 2}}} \right) - \mathop {\lim }\limits_{x \to 0} \log {x^2}}  \cr 
  & L = {L_1} - {L_2} = {e^{1 - \mathop {\lim }\limits_{x \to 0} \log {x^2}}} - {e^{1 - \mathop {\lim }\limits_{x \to 0} \log {x^2} + \mathop {\lim }\limits_{x \to 0} \log \left( {1 - {\textstyle{x \over 2}}} \right)}}  \cr 
  & \,\,\,\,\, = \,\,{e^{1 - \mathop {\lim }\limits_{x \to 0} \log {x^2}}}.\left( {1 - {e^{\mathop {\lim }\limits_{x \to 0} \log \left( {1 - {\textstyle{x \over 2}}} \right)}}} \right)  \cr 
  & \,\,\,\,\, = \,\,\mathop {\lim }\limits_{x \to 0} {e \over {{x^2}}}\left( {1 - \left( {1 - {\textstyle{x \over 2}}} \right)} \right)  \cr 
  & \,\,\,\,\, = \,\,\mathop {\lim }\limits_{x \to 0} \,{e \over {2x}} = \infty  \cr} $$
 A: In your very first step,  you cannot break a limit into the subtraction of two limits unless both of the other limits exist.  The theorem:
$\lim _{x\to a} (f(x)+g(x))=\lim _{x\to a}f(x) +\lim _{x\to a}g(x)$ is ONLY valid if the two limits on the right hand side exist.  In your case, the second limit clearly does not exist, because it goes to infinity.

Edit for clarity,  neither does the first limit. So in effect, what you tried to do was make this an $\infty - \infty$,  which doesn't work as seperate limits, but does work together (sometimes)
A: IF  $\lim f(x) = k$ and $\lim g(x) = j$ then then $\lim [ f(x)\pm g(x)] = \lim f(x) \pm \lim g(x)$ but ONLY if both limits of $f(x)$ exist and are finite.
we can take this a step further.  If $\lim f(x) = k$ but $\lim g(x)$ doesn't exist than $\lim [f(x) \pm g(x)]$ doesn't exist.   We can even take it so for that if $\lim f(x) = \infty$ and $\lim g(x) = \inf$ then $\lim [f(x) + g(x)]=\infty$
But if $\lim f(x)= \infty$ and $\lim g(y) = \infty$ (or either not exist) then $\lim [f(x) - g(x)]$ is not determinable (by this alone at any rate.)
An obvious counter example is $1= \lim 1 = \lim [(x+1) - x]\ne \lim (x+1) - \lim x \text{ air quotes "equals"-- wink-wink } \infty - \infty = \text{who the heck knows}$.
