- #1
tmt1
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If I have this sequence
$$a_n = \ln\left({\frac{n}{n^2 + 1}}\right)$$
I need to find:
$$ \lim_{{n}\to{\infty}} \ln\left({\frac{n}{n^2 + 1}}\right)$$
Shouldn't I be able to find the limit of$$ \lim_{{n}\to{\infty}} \frac{n}{n^2 + 1}$$ (which is $0$) and then substitute the result of that into the original limit and get the answer there?
So if I substitute in 0 I would get
$$ \lim_{{n}\to{\infty}} \ln\left({0}\right)$$
which would be negative $\infty$. However this is the incorrect answer.
$$a_n = \ln\left({\frac{n}{n^2 + 1}}\right)$$
I need to find:
$$ \lim_{{n}\to{\infty}} \ln\left({\frac{n}{n^2 + 1}}\right)$$
Shouldn't I be able to find the limit of$$ \lim_{{n}\to{\infty}} \frac{n}{n^2 + 1}$$ (which is $0$) and then substitute the result of that into the original limit and get the answer there?
So if I substitute in 0 I would get
$$ \lim_{{n}\to{\infty}} \ln\left({0}\right)$$
which would be negative $\infty$. However this is the incorrect answer.
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