lim_(x→0) ((e^(x+2x+3x+4x+.....+nx) −1)/x)=?


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

$\lim_{x \to 0} \frac{e^{x + 2 x + 3 x + 4 x + . . . . . + n x} - 1}{x} = ?$
	Answered by mehdee42 last updated on 18/Mar/23

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

	$e^{x + 2 x + 3 x + . . . + n x} = e^{\frac{n (n + 1) x}{2}} = 1 + \frac{n (n + 1) x}{2} + \frac{n^{2} {(n + 1)}^{2} x^{2}}{2^{2} 2!} + . . .$ $\lim_{x \to 0} \frac{e^{x + 2 x + 3 x + 4 x + . . . . . + n x} - 1}{x}$ $= \lim_{x \to 0} [\frac{n (n + 1)}{2} + \frac{n^{2} {(n + 1)}^{2} x}{2^{2} 2!} + . . .]$ $= \frac{n (n + 1)}{2}$
	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

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