lim-x-0-e-x-2x-3x-4x-nx-1-x-

Question Number 189501 by mathlove last updated on 18/Mar/23

\lim_{x \to 0} \frac{e^{x + 2 x + 3 x + 4 x + \dots . . + n x} - 1}{x} = ?

Answered by mehdee42 last updated on 18/Mar/23

h o p \to {l i m}_{x \to 0} (1 + 2 + \dots + n) e^{x + 2 x + \dots + n x} = \frac{n (n + 1)}{2}

Commented by mathlove last updated on 18/Mar/23

w i t h o u t H o p e t a l R u l

Answered by mr W last updated on 18/Mar/23

e^{x + 2 x + 3 x + \dots + n x} = e^{\frac{n (n + 1) x}{2}} = 1 + \frac{n (n + 1) x}{2} + \frac{n^{2} {(n + 1)}^{2} x^{2}}{2^{2} 2!} + \dots

\lim_{x \to 0} \frac{e^{x + 2 x + 3 x + 4 x + \dots . . + n x} - 1}{x}

= \lim_{x \to 0} [\frac{n (n + 1)}{2} + \frac{n^{2} {(n + 1)}^{2} x}{2^{2} 2!} + \dots]

= \frac{n (n + 1)}{2}

Commented by mathlove last updated on 18/Mar/23

t h a n k s d e a r

Answered by CElcedricjunior last updated on 18/Mar/23

l = \frac{n (n + 1)}{2}