# another exponential limit

• June 19th 2008, 03:51 PM
winterwyrm
another exponential limit
Thanks so much! The limit as n--> infinity of [1 + 7/n + 1/(n^4)]^n
• June 19th 2008, 04:00 PM
Moo
Hello,

Quote:

Originally Posted by winterwyrm
Thanks so much! The limit as n--> infinity of [1 + 7/n + 1/(n^4)]^n

$(1+\frac 7n+\frac 1{n^4})^n$

Taking the logarithm :

$n \ln(1+\frac 7n+\frac 1{n^4})$

We know that $\lim_{x \to 0} \ln(1+x)=\lim_{x \to 0} x$

Here, $\lim_{n \to \infty} \frac 7n+\frac 1{n^4}=0$.

So $\lim_{n \to \infty} n \ln(1+\frac 7n+\frac 1{n^4})=\lim_{n \to \infty} n( \frac 7n+\frac 1{n^4})=\lim_{n \to \infty} 7+\frac 1{n^3}=7$

By taking back the exponential (because we took the logarithm) :

$\lim_{n \to \infty}(1+\frac 7n+\frac 1{n^4})^n=\lim_{n \to \infty} e^{n \ln(1+\frac 7n+\frac 1{n^4})}=e^7$
• June 19th 2008, 04:48 PM
winterwyrm
Brighter than a thousand suns.