# How do I find $\lim_{n\to\infty}(\frac{n-1}{n})^n$

How would I find the limit for:

$$\lim_{n\to\infty}\left(\frac{n-1}{n}\right)^n$$

I know it approaches $\frac{1}{e}$, but I have no idea how it works. Plus, why does: $$\lim_{n\to\infty}\left(\frac{n-x}{n}\right)^n=\frac{1}{e^x}$$

-
– matt Mar 27 '12 at 9:35
How has $e$ been defined to you? – Américo Tavares Mar 27 '12 at 9:57
e = $\lim_{n\to\infty}{(\frac{n+1}{n})^n}$ – user27251 Mar 27 '12 at 10:00
Thanks for the information. And is $n$ an integer? I assume it is. Would you mind confirming? – Américo Tavares Mar 27 '12 at 10:13

1
@user27251 Hint: $\frac{n-1}n = \frac 1{\frac n{n-1}}$ – martini Mar 27 '12 at 10:56
show 2 more comments

## 5 Answers

1. (See martini's comment). For the first question write $\frac{n-1}{n}$ as $$\begin{equation*} \frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}. \end{equation*}$$ Apply limits and use the definition of $\mathrm{e}$ you have been given $$\begin{equation*} \mathrm{e}=\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n}. \end{equation*}$$ We have $$\begin{eqnarray*} \lim_{n\rightarrow \infty }\left( \frac{n-1}{n}\right) ^{n} &=&\lim_{n\rightarrow \infty }\frac{1}{\left( \frac{n}{n-1}\right) ^{n}}= \frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n}{n-1}\right) ^{n}} \\ &=&\frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n+1}}= \frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n}\cdot\lim_{n\rightarrow \infty }\frac{n+1}{n}} \\ &=&\frac{1}{\mathrm{e}\cdot 1}=\frac{1}{\mathrm{e}}. \end{eqnarray*}$$
2. Hint for the second question: write $$\begin{equation*} \left( \frac{n-x}{n}\right) ^{n}=\left( \left( 1-\frac{1}{n/x}\right) ^{n/x}\right) ^{x}, \end{equation*}$$ use the substitution $m=n/x$ and apply limits.
-

One of the basic properties of $e$ is that $$\lim_{ n \to \infty} \left(1+\frac{1}{n} \right)^n=e$$

You can use this here to find your answer.

-

For the first, take the reciprocal and substitute $n\mapsto n+1$ to get $$\begin{align} \frac{1}{\lim\limits_{n\to\infty}\left(\frac{n-1}{n}\right)^n} &=\lim_{n\to\infty}\left(\frac{n}{n-1}\right)^n\\ &\stackrel{n\to n+1}{=}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n+1}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\\ &=e\cdot1 \end{align}$$ Therefore, $\lim\limits_{n\to\infty}\left(\frac{n-1}{n}\right)^n=\dfrac1e$.

For the second, take the reciprocal and substitute $n\mapsto nx+x$ to get $$\begin{align} \frac{1}{\lim\limits_{n\to\infty}\left(\frac{n-x}{n}\right)^n} &=\lim_{n\to\infty}\left(\frac{n}{n-x}\right)^n\\ &\stackrel{n\to nx+x}{=}\lim_{n\to\infty}\left(\frac{nx+x}{nx}\right)^{nx+x}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^x\\ &=e^x\cdot1 \end{align}$$ Therefore, $\lim\limits_{n\to\infty}\left(\frac{n-x}{n}\right)^n=\dfrac{1}{e^x}$.

We could have started from the second and set $x=1$ for the first.

-

I do the first part and the second part is done similarly.

Solution 1 We know that the derivative of $\ln x$ is $\frac{1}{x}$. Hence using first principle differentiation, we have: $$\begin{align*} \frac{1}{x}&=\lim_{h\to0}\frac{\ln(x+h)-\ln x}{h}\\ &=\lim_{h\to0}\ln\left(1+\frac{h}{x}\right)^{\frac{1}{h}} \end{align*}$$ Let $n=\frac{1}{h}$ and $y=\frac{1}{x}$, so now $$y=\lim_{n\to\infty}\ln\left(1+\frac{y}{n}\right)^n$$ Now let $y=-1$, then $$\begin{align*} -1&=\lim_{n\to\infty}\ln\left(1-\frac{1}{n}\right)^n\\ \frac{1}{e}&=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n \end{align*}$$

Solution 2 Use L'Hopital's rule: $$\begin{align*} \lim_{n\to\infty}n\ln \left(1-\frac{1}{n}\right)&=-\lim_{n\to\infty}\frac{1}{1-\frac{1}{n}}\\ &=-1 \end{align*}$$ The rest is like solution 1.

-
ln is not additive. – hilbert Mar 27 '12 at 14:50

2
What does that have to do with my solution? – Vafa Khalighi Mar 27 '12 at 23:47

Here's a good strategy, first do the limit of the log of expression.

Let $y = \left(1 - \frac{x}{n} \right)^n$ (which is the same as your expression). Then, take the natural log of both sides to get $$\ln y = \ln \left[\left(1 - \frac{x}{n} \right)^n \right] = n \cdot \ln \left(1 - \frac{x}{n} \right) = \frac{\ln \left(1 - \frac{x}{n} \right)}{\frac{1}{n}}$$ Now, as $n \to \infty$, the final fraction goes to $\frac{0}{0}$, an indeterminate form. This suggests that we try l'Hopital's rule. That is

$$\begin{align*} \lim_{n \to \infty} \ln y &= \lim_{n \to \infty} \frac{\ln \left(1 - \frac{x}{n} \right)}{\frac{1}{n}} \\ &= \lim_{n \to \infty} \frac{\left(1 - \frac{x}{n}\right)^{-1} (x \cdot n^{-2})}{-n^{-2}} \\ &= \lim_{n \to \infty} -x\left(1 - \frac{x}{n} \right)^{-1} \\ &= -x \end{align*}$$

Therefore, since the exponential function $\exp(x) = e^x$ is continuous, we can move the limit in or outside of this function (by the definition of continuity) and thus find the limit of $y$ itself:

$$\lim_{n \to \infty} y = \lim_{n \to \infty} \exp(\ln y) = \exp \left( \lim_{n \to \infty} \ln y \right) = \exp (-x) = e^{-x}$$

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 16, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.909809410572052, "perplexity_flag": "head"}