# Thread: Finding limits w/definition not shortcuts

1. ## Finding limits w/definition not shortcuts

My question is how to get the limit of lim ((3n+1)/(n+2))=3.

I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.

However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.

Any suggestions?

2. ## Re: Finding limits w/definition not shortcuts

Originally Posted by CountingPenguins
My question is how to get the limit of lim ((3n+1)/(n+2))=3.

I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.
I'm sort of confused by your post. So are you suppose to prove that $\displaystyle \lim\frac{3n+1}{n+2}=3$?

If yes, then start with the inequality $\displaystyle |\frac{3n+1}{n+2}-3|=|\frac{3n+1}{n+2}-\frac{3(n+2)}{n+2}|<|\frac{-5}{n}|$.

3. ## Re: Finding limits w/definition not shortcuts

Originally Posted by CountingPenguins
Any suggestions?
Some more details: $\displaystyle \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$ now, for every $\displaystyle \epsilon>0$ choose $\displaystyle N>\dfrac{5}{ \epsilon}-2$ .

4. ## Re: Finding limits w/definition not shortcuts

[QUOTE=FernandoRevilla;662680]Some more details: $\displaystyle \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$

Thank you for your response, but can you tell me how you got from $\displaystyle \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$ it isn't obvious to me. Those are the details I am missing.

Thanks,
CP

5. ## Re: Finding limits w/definition not shortcuts

$\displaystyle \left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \left |\dfrac{3n+1-3n-6}{n+2}\right |< \epsilon\Leftrightarrow$

$\displaystyle \left |\dfrac{-5}{n+2}\right |< \epsilon\Leftrightarrow \dfrac{5}{n+2}< \epsilon\Leftrightarrow\ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$

6. ## Re: Finding limits w/definition not shortcuts

Originally Posted by CountingPenguins
My question is how to get the limit of lim ((3n+1)/(n+2))=3.

I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.

However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.

Any suggestions?
To prove $\displaystyle \displaystyle \lim_{x \to \infty}f(x) = L$, you need to show that $\displaystyle \displaystyle x > N \implies |f(x) - L| < \epsilon$ for $\displaystyle \displaystyle \epsilon > 0$. So in this case, to prove $\displaystyle \displaystyle \lim_{n \to \infty}\frac{3n + 1}{n + 2} = 3$, you need to show $\displaystyle \displaystyle n > N \implies \left|\frac{3n + 1}{n + 2} - 3\right| < \epsilon$.

\displaystyle \displaystyle \begin{align*} \left|\frac{3n + 1}{n + 2} - 3\right| &< \epsilon \\ \left|\frac{3n + 1 - 3(n + 2)}{n + 2}\right| &< \epsilon \\ \left|-\frac{5}{n+2}\right| &< \epsilon \\ \frac{5}{|n + 2|} &< \epsilon \\ \frac{|n + 2|}{5} &> \frac{1}{\epsilon} \\ |n + 2| &> \frac{5}{\epsilon} \\ n + 2 &> \frac{5}{\epsilon}\textrm{ since }\epsilon > 0\textrm{ and }n > 0 \\ n &> \frac{5}{\epsilon} - 2\end{align*}

So by letting $\displaystyle \displaystyle N = \frac{5}{\epsilon} - 2$, the proof will follow.

7. ## Re: Finding limits w/definition not shortcuts

Thank you very much for filling in all the blanks for me. I don't want the answer so much as how and why it works and you showed me how. Thanks.