Limit of (1+1/n)^n as n approaches infinity

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

Limit of (a^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:

limx→0 sin(√x)/x	limx→0 sin(x2)/x
limx→∞ sinx/x	limx→0 x/sinx

Limit of sin(1/x) when x→0

Limit of sin3x/sin2x when x→0

Limit of tanx/x when x→0

Limit of (cosx-1)/x when x→0

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