극한 계산기

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이 무료 계산기는 주어진 점(무한대 포함)에서 주어진 함수의 극한(양측 또는 편측, 좌극한과 우극한 포함)을 단계별 풀이와 함께 구해 줍니다.

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Solution

Your input: find $$$\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{n}$$$

Since `u=e^(ln(u))`, then:

$${\color{red}{\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{n}}} = {\color{red}{\lim_{n \to \infty} e^{\ln{\left(\left(1 + \frac{3}{n}\right)^{n} \right)}}}}$$

Simplify:

$${\color{red}{\lim_{n \to \infty} e^{\ln{\left(\left(1 + \frac{3}{n}\right)^{n} \right)}}}} = {\color{red}{\lim_{n \to \infty} e^{n \ln{\left(1 + \frac{3}{n} \right)}}}}$$

Move the limit under the exponential:

$${\color{red}{\lim_{n \to \infty} e^{n \ln{\left(1 + \frac{3}{n} \right)}}}} = {\color{red}{e^{\lim_{n \to \infty} n \ln{\left(1 + \frac{3}{n} \right)}}}}$$

Rewrite:

$$e^{{\color{red}{\lim_{n \to \infty} n \ln{\left(1 + \frac{3}{n} \right)}}}} = e^{{\color{red}{\lim_{n \to \infty} \frac{\ln{\left(1 + \frac{3}{n} \right)}}{\frac{1}{n}}}}}$$

Since we have an indeterminate form of type $$$\frac{0}{0}$$$, we can apply the l'Hopital's rule:

$$e^{{\color{red}{\lim_{n \to \infty} \frac{\ln{\left(1 + \frac{3}{n} \right)}}{\frac{1}{n}}}}} = e^{{\color{red}{\lim_{n \to \infty} \frac{\frac{d}{dn}\left(\ln{\left(1 + \frac{3}{n} \right)}\right)}{\frac{d}{dn}\left(\frac{1}{n}\right)}}}}$$

For steps, see derivative calculator.

$$e^{{\color{red}{\lim_{n \to \infty} \frac{\frac{d}{dn}\left(\ln{\left(1 + \frac{3}{n} \right)}\right)}{\frac{d}{dn}\left(\frac{1}{n}\right)}}}} = e^{{\color{red}{\lim_{n \to \infty} \frac{3}{1 + \frac{3}{n}}}}}$$

Simplify:

$$e^{{\color{red}{\lim_{n \to \infty} \frac{3}{1 + \frac{3}{n}}}}} = e^{{\color{red}{\lim_{n \to \infty} \frac{3 n}{n + 3}}}}$$

Apply the constant multiple rule $$$\lim_{n \to \infty} c f{\left(n \right)} = c \lim_{n \to \infty} f{\left(n \right)}$$$ with $$$c=3$$$ and $$$f{\left(n \right)} = \frac{n}{n + 3}$$$:

$$e^{{\color{red}{\lim_{n \to \infty} \frac{3 n}{n + 3}}}} = e^{{\color{red}{\left(3 \lim_{n \to \infty} \frac{n}{n + 3}\right)}}}$$

Multiply and divide by $$$n$$$:

$$e^{3 {\color{red}{\lim_{n \to \infty} \frac{n}{n + 3}}}} = e^{3 {\color{red}{\lim_{n \to \infty} \frac{n}{n \frac{n + 3}{n}}}}}$$

Divide:

$$e^{3 {\color{red}{\lim_{n \to \infty} \frac{n}{n \frac{n + 3}{n}}}}} = e^{3 {\color{red}{\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n}}}}}$$

The limit of the quotient is the quotient of limits:

$$e^{3 {\color{red}{\lim_{n \to \infty} \frac{1}{1 + \frac{3}{n}}}}} = e^{3 {\color{red}{\frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty}\left(1 + \frac{3}{n}\right)}}}}$$

The limit of a constant is equal to the constant:

$$e^{\frac{3 {\color{red}{\lim_{n \to \infty} 1}}}{\lim_{n \to \infty}\left(1 + \frac{3}{n}\right)}} = e^{\frac{3 {\color{red}{1}}}{\lim_{n \to \infty}\left(1 + \frac{3}{n}\right)}}$$

The limit of a sum/difference is the sum/difference of limits:

$$e^{3 {\color{red}{\lim_{n \to \infty}\left(1 + \frac{3}{n}\right)}}^{-1}} = e^{3 {\color{red}{\left(\lim_{n \to \infty} 1 + \lim_{n \to \infty} \frac{3}{n}\right)}}^{-1}}$$

The limit of a constant is equal to the constant:

$$e^{3 \left(\lim_{n \to \infty} \frac{3}{n} + {\color{red}{\lim_{n \to \infty} 1}}\right)^{-1}} = e^{3 \left(\lim_{n \to \infty} \frac{3}{n} + {\color{red}{1}}\right)^{-1}}$$

Apply the constant multiple rule $$$\lim_{n \to \infty} c f{\left(n \right)} = c \lim_{n \to \infty} f{\left(n \right)}$$$ with $$$c=3$$$ and $$$f{\left(n \right)} = \frac{1}{n}$$$:

$$e^{3 \left(1 + {\color{red}{\lim_{n \to \infty} \frac{3}{n}}}\right)^{-1}} = e^{3 \left(1 + {\color{red}{\left(3 \lim_{n \to \infty} \frac{1}{n}\right)}}\right)^{-1}}$$

The limit of a quotient is the quotient of limits:

$$e^{3 \left(1 + 3 {\color{red}{\lim_{n \to \infty} \frac{1}{n}}}\right)^{-1}} = e^{3 \left(1 + 3 {\color{red}{\frac{\lim_{n \to \infty} 1}{\lim_{n \to \infty} n}}}\right)^{-1}}$$

The limit of a constant is equal to the constant:

$$e^{3 \left(1 + \frac{3 {\color{red}{\lim_{n \to \infty} 1}}}{\lim_{n \to \infty} n}\right)^{-1}} = e^{3 \left(1 + \frac{3 {\color{red}{1}}}{\lim_{n \to \infty} n}\right)^{-1}}$$

Constant divided by a very big number equals $$$0$$$:

$$e^{3 \left(1 + 3 {\color{red}{1 \frac{1}{\lim_{n \to \infty} n}}}\right)^{-1}} = e^{3 \left(1 + 3 {\color{red}{\left(0\right)}}\right)^{-1}}$$

Therefore,

$$\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{n} = e^{3}$$

Answer: $$$\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^{n}=e^{3}$$$