Limit $\lim_{x \to \infty}\left (1- \frac1x\right)^x = ?$

Question

$$\lim_{x \to \infty} \left(1+ \frac1x\right)^x = e$$ , then$$\lim_{x \to \infty} \left(1- \frac1x\right)^x =\; ?$$

I tried $\lim_{x \to \infty} \left(1- \frac1x\right)^x = \lim_{x \to \infty}\left(\frac{x-1}{x}\right)^x$, but this is not helpful.

Note that from Bernoulli's Inequality we have for $x>0$$$1\ge\left(1+\frac1x\right)^x\left(1-\frac1x\right)^x=\left(1-\frac1{x^2}\right)^x\ge 1-\frac1x$$Now apply the squeeze theorem and conclude. — Mark Viola, May 22 '18 at 02:41
You might have a look at: limit of $\left( 1-\frac{1}{n}\right)^{n}$ or Finding the limit of $\left(\frac{n}{n+1}\right)^n$. You can probably find osme other posts about this limit - for example, you can search in Aproach0. — Martin Sleziak, May 22 '18 at 04:32

score 5 · Accepted Answer · answered May 22 '18 at 02:52

5

Try this:

$$ 1-\frac{1}{x}=\frac{1}{\frac{x}{x-1}}=\frac{1}{1+\frac{1}{x-1}} $$

so

$$ \lim_{x\rightarrow \infty} \left(1-\frac{1}{x}\right)^x=\lim_{x\rightarrow \infty} \left(\frac{1}{1+\frac{1}{x-1}}\right)^x=\lim_{x\rightarrow \infty} \frac{1}{\left(1+\frac{1}{x-1}\right)^{x-1}} \left(\frac{1}{1+\frac{1}{x-1}} \right)=\frac{1}{e}\cdot 1=\frac{1}{e} $$

answered May 22 '18 at 02:52

JasonM

3,151

That is a lot more work than one needs. Have a look at the comment I left under the OP. – Mark Viola May 22 '18 at 03:11
@MarkViola That is true, but I was just trying to give a different approach that perhaps used previously established facts. I understand why your last inequality holds, but perhaps the OP didn't – JasonM May 22 '18 at 04:13

score 2 · Answer 2 · answered May 22 '18 at 02:33

2

Let, $$y=(1-\frac{1}{x})^x$$

$$or, \ln y=x\ln(1-\frac{1}{x})$$ $$or,\ln y=\frac{\ln(1-\frac{1}{x})}{x^{-1}}$$

Now, $$or,\displaystyle\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}\frac{\ln(1-\frac{1}{x})}{x^{-1}}$$

Using L'Hospital rule, $$\displaystyle\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}-\frac{1}{1-\frac{1}{x}}=-1$$ So, $y=e^{-1}$.

answered May 22 '18 at 02:33

Sujit Bhattacharyya

3,030

The use of logarithms is completely unnecessary. – Mark Viola May 22 '18 at 03:11

score 1 · Answer 3 · answered May 22 '18 at 02:48

1

$$\lim_{x \to \infty} (1- \frac1x)^x =e^{-1} $$ is evaluated by taking logarithm of both sides of $ y=(1- \frac1x)^x $ and applying the L'Hospital Rule.

answered May 22 '18 at 02:48

Mohammad Riazi-Kermani

68,728

score 0 · Answer 4 · answered May 22 '18 at 03:13

Apply the common limit:$\lim _{ x\rightarrow \infty }{ (1+\frac { k }{ x } )^ x } =e^ k$

In this case we have,$$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}$$ So we compare it with $\lim _{ x\rightarrow \infty }{ (1+\frac { k }{ x } )^ x } =e^ k$ we have, $k=-1$ and $x=x$

So, $$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}=e^(-1)$$ $$\lim _{ x\rightarrow \infty }{ (1+(-\frac { 1 }{ x } )^ x )}=\frac1e$$

score 0 · Answer 5 · answered May 22 '18 at 03:21

Just another way.

$$y=\left(1-\frac{1}{x}\right)^x\implies \log(y)=x \log\left(1-\frac{1}{x}\right)$$ Now, using Taylor series, $$\log\left(1-\frac{1}{x}\right)=-\frac{1}{x}-\frac{1}{2 x^2}+O\left(\frac{1}{x^3}\right)$$ $$\log(y)=-1-\frac{1}{2 x}+O\left(\frac{1}{x^2}\right)$$ Continuing with Taylor series $$y=e^{\log(y)}=\frac{1}{e}-\frac{1}{2 e x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.

Even for "rather small" values of $x$, the approximation is quite good. For example, for $x=10$, the exact value is $3486784401 \times 10^{-10}=0.3486784401$ while the above formula gives $\frac{19}{20 e}\approx 0.349485$.

Limit $\lim_{x \to \infty}\left (1- \frac1x\right)^x = ?$

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