Limit involving the exponential of an expression plus a small-o

Since you've solved it using Qiaochu's logarithm hint, let's try another way. I am guessing that you already know that $$ \lim_{x\to\infty}\left[1+\frac{a}{x}\right]^x=e^a\tag{1} $$ and that the new twist is the little-o term. Little-o here means that for any $\epsilon>0$, there is an $M>0$ so that for $x>M$ $$ \left|o\left(\frac{1}{x}\right)\right|\le\frac{\epsilon}{x}\tag{2} $$ Condition $(2)$ means that for $x>M$ $$ \left[1+\frac{a-\epsilon}{x}\right]^x\le\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le\left[1+\frac{a+\epsilon}{x}\right]^x\tag{3} $$ Inequalities $(3)$ imply that $$ e^{a-\epsilon}\le\liminf_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le\limsup_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x\le e^{a+\epsilon}\tag{4} $$ and since inequalities $(4)$ hold for any $\epsilon>0$, we have that $$ \lim_{x\to\infty}\left[1+\frac{a}{x}+o\left(\frac{1}{x}\right)\right]^x=e^a $$ Note that we cannot apply the Squeeze Theorem to $(3)$ because the limits of the upper and lower bounds are not equal. Since $\epsilon>0$, $e^{a-\epsilon}\not=e^{a+\epsilon}$; however, since $\epsilon$ can be chosen arbitrarily small, $(4)$ allows us to squeeze the $\liminf$ and $\limsup$ together to produce a limit.