lim-x-0-1-x-1-x-e-x-

Question Number 80417 by jagoll last updated on 03/Feb/20

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

Commented by MJS last updated on 03/Feb/20

I tried to approximate and got something

very close to - \frac{e}{2} but I cannot prove the

exact value

Commented by mr W last updated on 03/Feb/20

{(1 + x)}^{\frac{1}{x}} = e (1 - \frac{x}{2} + \frac{11 x^{2}}{24} - \dots)

{(1 + x)}^{\frac{1}{x}} - e = e (- \frac{x}{2} + \frac{11 x^{2}}{24} - \dots)

\frac{{(1 + x)}^{\frac{1}{x}} - e}{x} = e (- \frac{1}{2} + \frac{11 x}{24} - \dots) \to - \frac{e}{2}

Commented by jagoll last updated on 03/Feb/20

v i a T a y l o r s e r i e s s i r ?

Commented by mr W last updated on 03/Feb/20

y e s!

Commented by jagoll last updated on 03/Feb/20

t h a n k y o u s i r

Facebook Tweet Pin