find the value of $\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}$ find the value of $$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}$$I use hospital law and can't find answer

A little different (again no L'Hopital): Set $g(x) = [\ln (1+x)]/x = 1-x/2+x^2/3 - \cdots .$ Our expression can be written
$$-\frac{e^{g(x)}-e^{g(0)}}{x-0} \to -(e^g)'(0) = -e^{g(0)}g'(0)= -e\cdot (-1/2) = e/2.$$
