prove $\lim\limits_{x \to \infty}(1+ 1/n)^n = e$ without using l'Hospitals rule

Question

I did $y=(1+1/n)^n)$ so $ln(y)=n*ln(1+1/n)$ so then $ln(y)=Ln(1+1/n)/(1/n)$ which sith l'hospitals rule becomes $ln(y)=(-1/(n^2 + n))/(-1/n^2)$ which simplifies to $n/(n+1)$ which using l'hospitals rule is $1$ so the answer is e. how do you solve this without l'hospitals rule

Answer 1

Using the dominated convergence theorem,

$$1=\int_0^1 1\,dx=\lim_{n\to\infty}\int_0^1\frac{n}{n+x}\,dx=\lim_{n\to\infty}\left(n\ln(n+1)-n\ln(n)\right)=\lim_{n\to\infty}\ln\left(\left(1+\frac{1}{n}\right)^n\right).$$

Taking $\exp$ of both sides and using the continuity of the exponential function, we get $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.

Answer 2

Let $f(x) = \ln(x)$ and then go through the process of evaluating $f'(1)$ using the formal definition of the derivative, noting ahead of time that the actual value of $f'(1)$ is $1$:

$$ 1 = f'(1) = \lim_{h \to 0}\frac{f(1+h)-f(1)}{h} = \lim_{x \to 0}\frac{f(1+x)-f(1)}{x} = \lim_{x \to 0}\ln(1 + x)^{1/x}; $$

and so,

$$ e^{1} = e^{\lim_{x \to 0}\ln(1+x)^{1/x}} = \lim_{x \to 0}(1+x)^{1/x}. \tag{$*$}$$

Now, let $n = \frac{1}{x}$.

Then substituting $n$ into $(*)$ and noting that as $x \to 0^{+}$ we have $n \to \infty$, it follows that

$$\lim_{n \to \infty}\left( 1 + \frac{1}{n} \right)^{n} = e.$$

Answer 3

Series expansions. $P_n=(1+\frac{1}{n})^n=\sum_{k=0}^n\binom{n}{k}(\frac{1}{n})^k$

$e=\sum_{k=0}^\infty\frac{1}{k!}$

For fixed $k$, $\binom{n}{k}(\frac{1}{n})^k=\frac{n(n-1)...(n-k+1)}{n^k}\frac{1}{k!}\to \frac{1}{k!}$ as $n\to \infty$

Since all the terms of $P_n$ are positive and less than corresponding terms of $e$, the sequence $P_n \to e$

Answer 4

This can be done straight from the definition of $\ln$ as $\ln(x) \equiv \int_{1}^x dt/t$ and $\exp$ as its inverse function. Then we have \begin{multline} \ln((1+1/n)^n) = \int_1^{(1+1/n)^n} \frac{dt}{t} = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t} + \int_{(1+1/n)^{n-1}}^{(1+1/n)^n} \frac{dt}{t} \\ = \int_1^{(1+1/n)^{n-1}} \frac{dt}{t} + \int_1^{1+1/n} \frac{du}{u} = \ln((1+1/n)^{n-1}) + \int_1^{1+1/n} \frac{du}{u}. \end{multline} Repeating this $n$ times gives $$ \ln((1+1/n)^n) = n\int_1^{1+1/n} \frac{du}{u}. $$ It's not hard to show from the properties of the integral that $$ \lim_{n\rightarrow\infty} n\int_1^{1+1/n} \frac{du}{u} = 1, $$ and that $\ln$ is continuous, so it is valid to say $$ \lim_{n\rightarrow\infty}\ln((1+1/n)^n) = \ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n.\right) $$ Thus, $$ \ln\left(\lim_{n\rightarrow\infty}(1+1/n)^n\right) = 1 $$ and so we have $$ \lim_{n\rightarrow\infty}(1+1/n)^n = \exp(1) \equiv e $$