Use the formal definition of a limit to prove that each sequence converges, and find
its limit:
$\frac{3n-11}{2n^2+n}$
#Prove convergence and find limit using formal definition
jolly frost
viral dockBOT
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Hay
jolly frost
i can reduce it to say it behaves like 1/n but im not sure if thats rigourous enough
cerulean imp
i can reduce it to say it behaves like 1/n but im not sure if thats rigourous en...
Depends on which theorems you are allowed to use, if any.
At the core, you just need to solve for $n$ in $\frac{3n-11}{2n^2+n}<\varepsilon$
plain oarBOT
SWR
jolly frost
At the core, you just need to solve for $n$ in $\frac{3n-11}{2n^2+n}<\varepsilon...
ya, the quadratic term makes things a little messy but ill give it a shot
cerulean imp
You can remove it with inequality tricks. That may be required if you can't define square roots yet
jolly frost
ok this might be a stupid question, but $|\frac{3n-11}{2n^2+n}|$ cant be represented as $\frac{3n-11}{2n^2+n}$ for all n in IN, so i cant just remove the absolute value function
plain oarBOT
Hay
jolly frost
we can define sqrts 👍
cerulean imp
ok this might be a stupid question, but $|\frac{3n-11}{2n^2+n}|$ cant be represe...
Make n sufficiently large
jolly frost
ah gotcha
jolly frost
@cerulean imp sorry for the ping, but would be correct to say:
$\epsilon 3 n^2 + \epsilon n > 3n - 11$ after some manipulation
$3 \epsilon n + \epsilon > 3 - \frac{11}{n} $ for sufficiently large n become
$3 \epsilon n + \epsilon > 3$ and then solve for n?
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Hay
