limx→0 (1+x)1/x−ex=

=limx→0 (1+x)1/x−ex Apply L-H rule then, =limx→0(1+x)1/xx1+x−ln(x+1)x21 =limx→0(1+x)1/xlimx→0x1+x−ln(x+1)x2 =elimx→0(1+x)−x(1+x)2−1(x+1)2x =elimx→01−(x+1)(x+1)2×2x =elimx→0-x(x+1)2 2x =-e2

limx→0 (1+x)1/x−ex=

=limx→0 (1+x)1/x−ex Apply L-H rule then, =limx→0(1+x)1/xx1+x−ln(x+1)x21 =limx→0(1+x)1/xlimx→0x1+x−ln(x+1)x2 =elimx→0(1+x)−x(1+x)2−1(x+1)2x =elimx→01−(x+1)(x+1)2×2x =elimx→0-x(x+1)2 2x =-e2

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

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$\frac{e}{2}$

$\frac{- e}{2}$

$\frac{2}{e}$

answer is C.

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Detailed Solution

$= lim_{x \to 0} \frac{(1 + x)^{1 / x} - e}{x} A p p l y L - H r u l e t h e n, = lim_{x \to 0} \frac{(1 + x)^{1 / x} [\frac{\frac{x}{1 + x} - \ln (x + 1)}{x^{2}}]}{1} = lim_{x \to 0} (1 + x)^{1 / x} lim_{x \to 0} [\frac{\frac{x}{1 + x} - \ln (x + 1)}{x^{2}}] = e lim_{x \to 0} \frac{(\frac{(1 + x) - x}{(1 + x)^{2}} - \frac{1}{(x + 1)})}{2 x} = e lim_{x \to 0} \frac{1 - (x + 1)}{(x + 1)^{2} \times 2 x} = e lim_{x \to 0} \frac{- x}{(x + 1)^{2} 2 x} = \frac{- e}{2}$