Finding limit when limit approches to infinity

Question

Finding $\displaystyle \lim_{y\rightarrow \infty}\bigg[\frac{y}{e}-y\bigg(\frac{y}{y+1}\bigg)^y\bigg]$

Attempt: Put $\displaystyle y=\frac{1}{z}$ . Then $\displaystyle \lim_{z\rightarrow 0}\bigg[\frac{1}{ze}-\frac{1}{z}\bigg(\frac{z}{z+1}\bigg)^{\frac{1}{z}}\bigg]$

Could some help me to solve it, thanks

That should be $\lim_{z\to 0}$. – Thomas Andrews Jun 22 '17 at 15:46 — Thomas Andrews, Jun 22 '17 at 15:46
Is the answer $\frac {-1}{2e} $ – Archis Welankar Jun 22 '17 at 15:54 — Archis Welankar, Jun 22 '17 at 15:54

Clayton · Answer 1 · 2017-06-24T23:40:54.703

2

Hint: $$\left(\frac{y}{y+1}\right)^y=\left(1-\frac1{y+1}\right)^{y+1-1}\longrightarrow\frac1e.$$

Update: Since the hint might be too difficult for some users: $$\lim_{y\to\infty}\left(1+\frac{-1}{y+1}\right)^{y+1}\cdot\underbrace{\left(1-\frac{1}{y+1}\right)^{-1}}_{\rightarrow 1}=\lim_{y\to\infty}\left(1+\frac{-1}{y}\right)^y=e^{-1}=\frac1e.$$

edited Jun 24 '17 at 23:40

answered Jun 22 '17 at 15:39

Clayton

24,751

That won't suffice here. – Mark Viola Jun 22 '17 at 15:47
@MarkViola: Do tell, because that is precisely how it can be solved without resorting to Taylor expansions... – Clayton Jun 24 '17 at 21:02
OK. I'll tell. You lucked out. You need to expand to second order to rigorously conclude. – Mark Viola Jun 25 '17 at 00:05

score 2 · Accepted Answer · answered Jun 22 '17 at 15:41

2

Consider $$\log\left[\left(\frac{y}{1+y}\right)^y\right] =-y\log\left(1+\frac1y\right)=-1+\frac1{2y}+O(y^{-2}).$$ Therefore $$\left(\frac{y}{1+y}\right)^y=e^{-1}\exp\left(\frac1{2y}+O(y^{-2}) \right)=e^{-1}\left(1+\frac1{2y}+O(y^{-2})\right)$$ etc.

answered Jun 22 '17 at 15:41

Angina Seng

158,341

1

It would add to the readability of your answer if you mention that in the last step you are using Taylor's expansion... – Miguel Jun 22 '17 at 18:32

Finding limit when limit approches to infinity

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