Does this sequence converge to the proposed limit?

[cragar]

Homework Statement

Verify, using the definition of convergence of a sequence, that
the following sequences converge to the proposed limit.
a) lim \frac{1}{6n^2+1}=0
b) lim \frac{3n+1}{2n+5}=\frac{3}{2}
c) lim \frac{2}{\sqrt{n+3}} = 0

The Attempt at a Solution

A sequence a_n converges to a real number a if for every ε there is
an N in the naturals such that whenever n≥N it follows that
|a_n-a|< \epsilon.
so for the first one I need \frac{1}{6n^2+1}< \epsilon
and then I turn it into \frac{1}{\epsilon}<6n^2+1
So i could pick an n large enough to make that happen.
on the second one I move the 3/2 over and then combine those
fractions with a common denominator and I get
|\frac{-12}{4n+10}|< \epsilon
Am I doing this right or am I way off.

 

[vela]

on the second one I move the 3/2 over and then combine those
fractions with a common denominator and I get
|\frac{-12}{4n+10}|< \epsilon
Am I doing this right or am I way off.

Good start, though the -12 should be -13. Now what?

 

[cragar]

so I have |\frac{-13}{4n+10}<\epsilon |
for any epsilon I can pick an n large enough to make that true.

 

[vela]

You should explicitly find what N will work.

 

[cragar]

that seems weird to me because \epsilon could be anything
so how would any fixed N work.

 

[vela]

You're given some $\epsilon > 0$, and then you have to find some N for that given $\epsilon$ such that the implication holds, so N will generally depend on $\epsilon$.

For example, if you had
$$\frac{1}{n} < \epsilon \qquad \Rightarrow \qquad n > \frac{1}{\epsilon},$$ you could choose N to be any integer greater than $1/\epsilon$.

 

[cragar]

so then I just solve for n in terms of \epsilon
.25(\frac{-13}{\epsilon }-10)&lt;n
thanks for your help by the way

 

[vela]

Since you're working with inequalities, you can do some simplifications to make the algebra less tedious:
$$\left|\frac{-13}{4n+10}\right| < \frac{16}{4n} = \frac{4}{n} < \epsilon$$So instead of that complicated expression you have, you can choose N to be an integer greater than 4/ε.