Question

could someone explain to me why this doesn't diverge?

Answer 1

It converges to \( e^{-4}\)

Answer 2

how were you able to figure this out? :/

Answer 3

let y = RHS

ln(y) = ln(RHS)

apply L'hospital rule

Answer 4

Let a be x real number. It can be shown that
\[

\lim_{n->\infty} \left(1+ \frac x n \right)^n=e^x
\] Here how to prove it
\[
y_n=\left(1+ \frac x n \right)^n\\
\ln(y_n) = n \ln\left(1+ \frac x n \right)=\frac{ \ln\left(1+ \frac x n \right) }{\frac 1 n}
\\
\text{ Apply L'Hospital rule now, taking derivative with respect to n}


\]

Answer 5

\[
\frac{\frac{\partial }{\partial n}\log
\left(\frac{x}{n}+1\right)}{\frac{\partial }{\partial
n}\frac{1}{n}}=-\frac{x}{\frac{(-1) \left(n^2
\left(\frac{x}{n}+1\right)\right)}{n^2}}=\frac{x}{\frac{x}{n}+1}

\]

Answer 6

\[
\lim_{n\to infty }\frac{x}{\frac{x}{n}+1}=x\\
\ln(y_n) \to x\\
y_n\to e^x
\]
We are done.