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How to get started on limit [n => infty] (1 + 7/n)^n
- Thread starter cheffy
- Start date
Hello, cheffy!
Galactus tossed you a BIG hint.
But if you don't know how to apply it, it's wasted, of course.
We are told that: \(\displaystyle \L\:\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)
Now we hammer the given problem into that form . . .
We have: \(\displaystyle \L\:\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^n \;=\;\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^n\right]^{\frac{7}{7}} \;=\;\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]^7\)
Note that if \(\displaystyle n\to\infty\), then \(\displaystyle \frac{n}{7} \to\infty\)
Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]^7 \;=\;\underbrace{\left[\lim_{\frac{n}{7}\to\infty}\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]}_{\text{This is e}} ^7\)
Therefore: \(\displaystyle \L\:\lim_{n\to\infty}\left(1\,+\,\frac{7}{n}\right)^n\;=\;e^7\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
In general: \(\displaystyle \L\:\lim_{n\to\infty}\left(1\,+\,\frac{k}{n}\right)^n \;=\;e^k\)
Galactus tossed you a BIG hint.
But if you don't know how to apply it, it's wasted, of course.
\(\displaystyle \L\lim_{n \to \infty} \left(1\,+\,\frac{7}{n}\right)^n\)
We are told that: \(\displaystyle \L\:\lim_{z\to\infty}\left(1\,+\,\frac{1}{z}\right)^z\;=\;e\)
Now we hammer the given problem into that form . . .
We have: \(\displaystyle \L\:\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^n \;=\;\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^n\right]^{\frac{7}{7}} \;=\;\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]^7\)
Note that if \(\displaystyle n\to\infty\), then \(\displaystyle \frac{n}{7} \to\infty\)
Take the limit: \(\displaystyle \L\:\lim_{n\to\infty}\left[\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]^7 \;=\;\underbrace{\left[\lim_{\frac{n}{7}\to\infty}\left(1\,+\,\frac{1}{\left(\frac{n}{7}\right)}\right)^{\frac{n}{7}}\right]}_{\text{This is e}} ^7\)
Therefore: \(\displaystyle \L\:\lim_{n\to\infty}\left(1\,+\,\frac{7}{n}\right)^n\;=\;e^7\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
In general: \(\displaystyle \L\:\lim_{n\to\infty}\left(1\,+\,\frac{k}{n}\right)^n \;=\;e^k\)