$\displaystyle $ \lim_{x\to 0}$ $\displaystyle \frac{(1+x)^\frac{1}{x}-e+e\frac{x}{2}}{x^2}$
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Note that $\displaystyle \lim _{x \to 0} (1 + x)^{1/x} = \lim _{x \to \infty} (1 + \frac 1 x )^x = e $ So the prroblem becomes: $\displaystyle \lim _ {x \to 0} \frac {e - e + e \frac x 2 } {x^2} = \lim _{x \to 0} \frac e {2x} = \infty$.
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