Question

Tentukan limit fungsi euler dengan bentuk berikut
[tex]\displaystyle \lim_{n\to\infty}\sqrt[n]{\left ( 1+\frac{1}{2} \right )\left ( 1+\frac{1}{3} \right )\left ( 1+\frac{1}{4} \right )\left ( 1+\frac{1}{5} \right )...\left ( 1+\frac{1}{n} \right )}[/tex]

Answer 1

Verified answer

[tex]\begin{aligned}&\lim_{n\to\infty}\sqrt[n]{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right )\left(1+\frac{1}{5}\right){\dots}\left(1+\frac{1}{n}\right)}\\\vphantom{\Big|}&=\boxed{\,\bf1\,}\ .\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah:

Kita akan menentukan nilai dari

[tex]\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right )\left(1+\frac{1}{5}\right){\dots}\left(1+\frac{1}{n}\right)}\end{aligned}[/tex]

Kita perhatikan bahwa:

[tex]\begin{aligned}&\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right )\left(1+\frac{1}{5}\right){\dots}\left(1+\frac{1}{n}\right)\\&{=\ }\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\frac{6}{5}\cdot{\dots}\cdot\frac{n+1}{n}\\&{=\ }\frac{1}{2}\cdot\frac{\cancel{3\cdot4\cdot5\cdot6\cdot{\dots}\cdot n}\cdot(n+1)}{\cancel{3\cdot4\cdot5\cdot6\cdot{\dots}\cdot n}}\\&{=\ }\frac{n+1}{2}\end{aligned}[/tex]

Maka nilai limit yang kita cari adalah:

[tex]\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n+1}{2}}&=\lim_{n\to\infty}\frac{\sqrt[n]{n+1}}{\sqrt[n]{2}}\\&=\frac{\lim\limits_{n\to\infty}\sqrt[n]{n+1}}{\lim\limits_{n\to\infty}\sqrt[n]{2}}\\&=\frac{\lim\limits_{n\to\infty}\sqrt[n]{n+1}}{\lim\limits_{n\to\infty}2^{1/n}}\\&=\frac{\lim\limits_{n\to\infty}\sqrt[n]{n+1}}{2^{\lim\limits_{n\to\infty}(1/n)}}\\&=\frac{\lim\limits_{n\to\infty}\sqrt[n]{n+1}}{2^{0}}\\&=\frac{\lim\limits_{n\to\infty}\sqrt[n]{n+1}}{1}\end{aligned}[/tex]
[tex]\begin{aligned}&=\lim_{n\to\infty} \sqrt[n]{n+1}\\&=\lim_{n\to\infty} e \, {}^{\ln \left( \sqrt[n]{n+1} \right)}\\&=\lim_{n\to\infty} e \, {}^{\frac{\ln(n+1)}{n}}\\\lim_{n\to\infty} \sqrt[n]{\frac{n+1}{2}}& = e \, {}^{ \lim\limits_{n\to\infty} \frac{\ln(n+1)}{n} } \quad...(i)\end{aligned}[/tex]

Untuk nilai pangkatnya:

[tex]\begin{aligned}\lim_{n\to\infty}\frac{\ln(n+1)}{n}&=\lim_{n\to\infty}\frac{\frac{d}{dn}(\ln(n+1))}{\frac{d}{dn}(n)}\\&=\lim_{n\to\infty}\frac{\left(\dfrac{1}{n+1}\right)}{1}\\&=\lim_{n\to\infty}\frac{1}{n+1}\\&=\frac{\lim\limits_{n\to\infty}1}{\lim\limits_{n\to\infty}(n+1)}\\\lim_{n\to\infty}\frac{\ln(n+1)}{n}&=0\end{aligned}[/tex]

Maka, melanjutkan [tex](i)[/tex]:

[tex]\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n+1}{2}}&=e\,{}^{\lim\limits_{n\to\infty}\frac{\ln(n+1)}{n}}\\&=e^0\\&=\boxed{\bf1} \leftarrow \sf Jawaban\end{aligned}[/tex]

[tex]\blacksquare[/tex]