The limit of (1+1/n)^n as n approaches infinity is equal to e. The lim_{n→∞} (1+1/n)^{n} formula is given by

$\lim\limits_{n \to \infty} \big(1+\dfrac{1}{n} \big)^n=e$.

That is, the limit of (1+1/x)^{x} when x→∞ is equal to e. In this post, we will learn to prove the limit of log(1+x)/x when x tends to zero.

## Lim_{n→∞} (1 + 1/n)^{n}

**Question: **What is the limit of (1+1/n)^{n} when n tends to infinity?

Answer: The limit of (1+1/n)^{n} is equal to e when n tends to infinity. |

**Explanation:**

Let the given limit is equal to y, that is,

$y=\lim\limits_{n \to \infty} \big(1+\dfrac{1}{n} \big)^n$.

We need to find the value of y. Taking natural logarithm ln=log_{e} on both sides, we get that

$\ln y=\ln \lim\limits_{n \to \infty} \big(1+\dfrac{1}{n} \big)^n$.

⇒ $\ln y=\lim\limits_{n \to \infty} \ln \big(1+\dfrac{1}{n} \big)^n$ by the rule of lim_{x→∞} ln(f(x)) = ln lim_{x→∞} f(x).

⇒ $\ln y=\lim\limits_{n \to \infty}$ $\Big[ n \ln \big(1+\dfrac{1}{n} \big) \Big]$ using the logarithm rules

⇒ $\ln y=\lim\limits_{n \to \infty}$ $\Big[ n \big(\dfrac{1}{n}-\dfrac{1}{2n^2}+\dfrac{1}{3n^3}-\dfrac{1}{4n^4}+\cdots \big) \Big]$ using the formula of log(1+x).

⇒ $\ln y=\lim\limits_{n \to \infty}$ $\Big( 1-\dfrac{1}{2n}+\dfrac{1}{3n^2}-\dfrac{1}{4n^3}+\cdots \Big)$

⇒ $\ln y=1-0+0-0+\cdots $

⇒ $\ln y=1 =\ln e$

⇒ $y=e$

So the limit of (1+1/n)^n is equal to e when x→∞.

**Read Also:** Limit of (e^{x}-1)/x when x→0

**Remark:** If we put n=x in the above limit formula, then we will get the limit of (1+1/x)^{x} when x tends to infinity. Thus, the formula of limit of (1+1/x)^{x} is equal to e when x→∞.

More Limits:

## FAQs

### Q1: What is the limit of (1+1/n)^{n} when n tends to ∞?

Answer: The limit of (1+1/n)^{n} when n tends to ∞ is equal to e, that is, lim_{n→∞} (1+1/n)^{n} =e.

### Q2: What is the limit of (1+1/x)^{x} when x tends to ∞?

Answer: The limit of (1+1/x)^{x} when x tends to ∞ is equal to e, that is, lim_{x→∞} (1+1/x)^{x} =e.