# Thread: Rigorous Proof on Limits

1. ## Rigorous Proof on Limits

I'd someone to check my work:

Using the definition of the convergence of a sequence, verify that the following converge to the limit that is proposed.

NOTE, the definition of convergence for a sequence is as follows: A sequence (a_n) converges to $a \in \mathbb{R}$ if, for every pos. number epsilon, there exists an $N \in \mathbb{N}$ such that whenever n >= N, it follows that |a_n - a| < epsilon.

1.) lim(1/(6n^2 + 1)) = 0

2.) lim((3n+1)/(2n+5) = 3/2

3.) lim(2/sqrt(n+3)) = 0

MY WORK:

1.) 1/(6n^2 + 1) < 1/(6n^2) < epsilon => n > sqrt(1/(6*epsilon))

2.) |(3n+1)/(2n+5) - 3/2| = |[2*(3n+1) - 3*(2n+5)]/[2*(2n+5]| = 13/2 * 1/(2n + 5) < 13/4n < epsilon => n > 13/4*epsilon

3.) 2/sqrt(n+3) < 2/sqrt(n) < epsilon => n > (2/epsilon)^2

There we go.. my professor doesn't like that I didn't take it back and show |a_n - a| < epsilon or something like that.. not quite sure what he wants.

2. Originally Posted by seerTneerGevoLI
I'd someone to check my work:

Using the definition of the convergence of a sequence, verify that the following converge to the limit that is proposed.

NOTE, the definition of convergence for a sequence is as follows: A sequence (a_n) converges to $a \in \mathbb{R}$ if, for every pos. number epsilon, there exists an $N \in \mathbb{N}$ such that whenever n >= N, it follows that |a_n - a| < epsilon.

1.) lim(1/(6n^2 + 1)) = 0

2.) lim((3n+1)/(2n+5) = 3/2

3.) lim(2/sqrt(n+3)) = 0

MY WORK:

1.) 1/(6n^2 + 1) < 1/(6n^2) < epsilon => n > sqrt(1/(6*epsilon))

2.) |(3n+1)/(2n+5) - 3/2| = |[2*(3n+1) - 3*(2n+5)]/[2*(2n+5]| = 13/2 * 1/(2n + 5) < 13/4n < epsilon => n > 13/4*epsilon

3.) 2/sqrt(n+3) < 2/sqrt(n) < epsilon => n > (2/epsilon)^2

There we go.. my professor doesn't like that I didn't take it back and show |a_n - a| < epsilon or something like that.. not quite sure what he wants.
here's what he wants. i'll just do the first one.

Proof: $\bold { \lim_{n \to \infty} \frac {1}{6n^2 + 1} = 0 }$

Let $\epsilon > 0$, choose $N \in \mathbb {N}$ such that $N > \sqrt {\frac 1{6 \epsilon}}$.

Then, $n \ge N \implies \left| \frac 1{6n^2 + 1} - 0 \right| = \left| \frac 1{6n^2 + 1}\right| < \left| \frac 1{6n^2} \right| < \left| \frac 1{6N^2}\right| < \left| \frac 1{6 \left( \sqrt { \frac 1{6 \epsilon}} \right)^2 } \right| = \epsilon$

Thus, $\lim_{n \to \infty} \frac 1{6n^2 + 1} = 0$

QED

Do you see the difference between what I did and what you did? Which one of our methods do you think conforms to the definition of the limit of a sequence that was given?

3. Jhevon, can you look over these responses and tell me if these are fine?

1-
Let epsilon> 0 and let N be a natural number such that 1/(6N^2)< epsilon. Then if n>= N we will have:
|1/(6n^2+1)|< |1/(6n^2)|<= |1/(6N^2)|< epsilon
And that is what we were looking for.

2-
Let epsilon> 0 and let N be a natural number such that 2/sqrt(N)< epsilon. Then if n>= N we will have:
|2/sqrt(n+3)|< |2/sqrt(n)|<= |2/sqrt(N)|< epsilon
And that is what we were looking for.

3-
Let epsilon>0 and let N be a natural number such that 13/(4N)< epsilon. Then if n>= N we will have:
|(3n+1)/(2n+5)-3/2|= |-13/(4n+10)|= |13/(4n+10)|< |13/(4n)|<= |13/(4N)|< epsilon
And that is what we were looking for.

4. Originally Posted by seerTneerGevoLI
Jhevon, can you look over these responses and tell me if these are fine?

1-
Let epsilon> 0 and let N be a natural number such that 1/(6N^2)< epsilon. Then if n>= N we will have:
|1/(6n^2+1)|< |1/(6n^2)|<= |1/(6N^2)|< epsilon
And that is what we were looking for.

2-
Let epsilon> 0 and let N be a natural number such that 2/sqrt(N)< epsilon. Then if n>= N we will have:
|2/sqrt(n+3)|< |2/sqrt(n)|<= |2/sqrt(N)|< epsilon
And that is what we were looking for.

3-
Let epsilon>0 and let N be a natural number such that 13/(4N)< epsilon. Then if n>= N we will have:
|(3n+1)/(2n+5)-3/2|= |-13/(4n+10)|= |13/(4n+10)|< |13/(4n)|<= |13/(4N)|< epsilon
And that is what we were looking for.
it is not really my taste to do these proofs like that, but those are fine. i don't know if i like the line "And that is what we were looking for," but hey, to each his own.

why did you change the way you chose N?

My current post is 3,666... you see that? 666!