Showing that $\lim_{n \rightarrow \infty}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n = e$

Question

$\lim_{n \rightarrow \infty}\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n = e$

I got this limit while solving another problem. Usually limits that look like this can be easily handled by factoring expressions, but this limit has addition inside the parenthesis which confuses me. How can this limit be solved? No l'Hopitals, thank you :)

Answer 1

Well note $\frac{1}{n}+\frac{1}{n^2}=\frac{n+1}{n^2}$. And you know $\lim\limits_{n\to\infty}(1+\frac{1}{n})^{n}=e$ We now write your expression as $$\left(1+\frac{n+1}{n^2}\right)^{\frac{n^2}{n+1} \times \frac{n+1}{n}}=e^{\lim\limits_{n\to\infty}\frac{n+1}{n}}=e$$

Answer 2

$(1+\frac 1 n)^{n} \leq (1+\frac 1 n+\frac 1 {n^{2}})^{n}$. Now $(1+\frac 1 n+\frac 1 {n^{2}})^{n}=e^{n\log(1+\frac 1 n +\frac 1 {n^{2}})}$ and you can use the inequality $\log(1+x) \leq x$ ($x>0$) to get the upper bound $e^{1+\frac 1 n}$. Now use squeeze theorem to complete the proof.

Answer 3

I would take logarithms and consider $$\ln\left[\left(1+\frac1 n+\frac 1{n^2}\right)^n\right] =n\ln\left(1+\frac 1n+\frac 1{n^2}\right)=n\sum_{k=1}^\infty \frac{(-1)^k}{k}\left(\frac1 n+\frac 1{n^2}\right)^k.$$ In this sum, apart from the first term, the sum of the remaining terms is $O(1/n^2)$, so the sum is $$n\left(\frac 1n+\frac 1{n^2}\right)+nO(1/n)=1+O(1/n)\to1.$$ As its logarithm tends to $1$, the original expression tends to $e$.

Answer 4

For any $\varepsilon>0$, we have $$ 1+\frac1n<1+\frac1n+\frac1{n^2}\leq 1+\frac{1+\varepsilon}{n} $$ for all $n$ large enough, so $$ \left(1+\frac1n\right)^n<\left(1+\frac1n+\frac1{n^2}\right)^n< \left(1+\frac{1+\varepsilon}{n}\right)^n. $$ Taking limit $n\to\infty$, we have $$ e\leq\lim_{n\to\infty}\left(1+\frac1n+\frac1{n^2}\right)^n\leq e^{1+\varepsilon}. $$ So $$ \lim_{n\to\infty}\left(1+\frac1n+\frac1{n^2}\right)^n=e. $$

Answer 5

It's easy to think with the definition: $$\lim_{n \rightarrow \infty}{(1+ \frac{1}{n})^n}=\lim_{t \rightarrow 0}{(1+ t)^{\frac{1}{t}}}=e$$

Therefore , $$\lim_{n \rightarrow \infty}{(1+ \frac{1}{n}+\frac{1}{n^2})^n}=\lim_{n \rightarrow \infty}{(1+ \frac{n+1}{n^2})^n}=\lim_{n \rightarrow \infty}{(1+ \frac{n+1}{n^2})^{ \frac{n^2}{n+1}\cdot \frac{n+1}{n^2}\cdot n}}=e^1=e$$

Answer 6

Use the formula for $1^\infty$ form. $$\lim_{x\to a}f(x)^{g(x)}=\mathrm{e}^{\lim_{x\to a}(f(x)-1)g(x)}$$ such that $\lim_{x\to a}f(x)=1$ and $\lim_{x\to a}g(x)=\infty$. Now using the formula here gives, $$L=\mathrm{e}^{\lim_{n\to \infty}(1+1/n)^n}=\mathrm{e}$$

Answer 7

Also with the limit of the log, but doing some asymptotic analysis: $$n\ln\Bigl(1+\frac1n+\frac1{n^2}\Bigr)=n\ln\Bigl(1+\frac1n+o\bigl(\frac1n\bigr)\Bigr)=n\Bigl(\frac1n+o\bigl(\frac1n\bigr)\Bigr)=1+o(1). $$