L'Hopitaling this will put you in the L'Hopital

Calculus Level 4

$\large \lim_{x \to \infty} \frac{x^x}{(1+x)^x} = E$

Find $\left \lfloor 1000 E \right \rfloor$ .

The answer is 367.

2 solutions

Hobart Pao
Apr 9, 2017

$= \displaystyle \lim_{x \to + \infty}\left( \dfrac{x}{1+x}\right)^x$ $= \displaystyle \lim_{x \to +\infty}\left( \dfrac{1+x}{x} \right)^{-x}$ $= \displaystyle \lim_{x \to +\infty}\left( 1 + \dfrac{1}{x} \right)^{-x}$ $= \displaystyle \lim_{x \to +\infty}\left[ \left( 1 + \dfrac{1}{x} \right)^{x} \right]^{-1}$ $= \dfrac{1}{e} \approx 0,3678794412$ $= \left \lfloor 1000 \dfrac{1}{e} \right\rfloor = \boxed{367}$

Same here. FYI, something typo $0,36787$ instead of $0.36787$ :)

Naren Bhandari - 3 years, 7 months ago

I got the value $\frac{1000}{e}$ right, but I've failed to put it in the floor function(...) I gave the answer 368 instead of 367.

Pepper Mint - 3 years, 7 months ago

Ayon Ghosh
Dec 14, 2017

Consider ; $\displaystyle \lim_{x \to \infty}\left( \dfrac{x}{1+x}\right)^x$ $= \displaystyle \lim_{x \to \infty}\left( 1 - \dfrac{1}{1+x}\right)^x$ $= \displaystyle \lim_{x \to \infty}\left[ \left( 1 - \dfrac{1}{1 +x} \right)^{x+1} \right]^{\dfrac{x}{x+1}}$ $\boxed{Since},$

$\displaystyle \lim_{x \to \infty}\left( 1 - \dfrac{1}{1+x}\right)^{1+x} = e^{-1}$ And $\displaystyle \lim_{x \to \infty}( \dfrac{x}{1+x} ) = 1$

So the limit $E = e^{-1}$

L'Hopitaling this will put you in the L'Hopital

The answer is 367.

2 solutions

0 pending reports