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