limit

**ayushdadhwal** · May 20th 2012, 09:40 AM

$$ \lim_{x\to 0}$ $\frac{(1+x)^\frac{1}{x}-e+e\frac{x}{2}}{x^2}$

**ayushdadhwal** · May 21st 2012, 09:05 AM

any view
thanks in advance

**ebaines** · May 22nd 2012, 12:23 PM

Note that

$\lim _{x \to 0} (1 + x)^{1/x} = \lim _{x \to \infty} (1 + \frac 1 x )^x = e$

So the prroblem becomes: $\lim _ {x \to 0} \frac {e - e + e \frac x 2 } {x^2} = \lim _{x \to 0} \frac e {2x} = \infty$ .

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