Let's push the limits!

Let $f(x)=(1+x)^{\frac{1}{x}}\quad \forall x \in \mathbb{R}^+ \cup (0,-1) \\ f(0)=e$

Notice that $f$ is continuous everywhere and differentiable on it's domain except possibly at zero.

We shall first figure out the maclaurin series expansion of $f(x)$ .

Now, $f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x}=\lim_{x \to 0} f'(x)=-\frac{1}{2}e$

The second equality follows by l'hopital's rule . Now evaluating the last limit is left as an exercise.

Similarly, we get, after some more work, $f''(0)=\frac{11}{24} e$

Hence, $f(x)=e\left( 1-\frac{1}{2}x + \frac{11}{24}x^2+\cdots \right)$

We hence conclude that $\large \lim_{x \to 0} \dfrac{(1+x)^{\frac{1}{x}}-e+ \frac{1}{2} ex}{x^2}=\frac{11}{24}e\\\implies \large \lim_{x \to 0} \dfrac{(1+x)^{\frac{1}{x}}-e+ \frac{1}{2} ex}{x^2 (1+x)^{\frac{1}{x}}}=\frac{11}{24}\\\implies a=11 \quad b=24 \implies a+b=\boxed{35}$