How Do You Calculate the Limit of (2n-1)/(3n+1) as n Approaches Infinity?

[shan]

This isn't really a homework question so feel free to move is somewhere else. My teacher tried to explain this to me but I didn't understand it so I thought I'd try hearing another person's point of view.

The question is
lim n->infinity (2n-1)/(3n+1)

The working out I was showed is
= lim n->infinity (2n+2/3)/(3n+1) - (1 2/3)/(3n+1)
= 2/3 - 0 = 2/3

What I don't understand is how to work out that the lim n->infinity (2n+2/3)/(3n+1) is 2/3. We were shown rules for limits as n tends to infinity but I don't think I've come across the above situation before.

 

[jzq]

This isn't really a homework question so feel free to move is somewhere else. My teacher tried to explain this to me but I didn't understand it so I thought I'd try hearing another person's point of view.

The question is
lim n->infinity (2n-1)/(3n+1)

The working out I was showed is
= lim n->infinity (2n+2/3)/(3n+1) - (1 2/3)/(3n+1)
= 2/3 - 0 = 2/3

What I don't understand is how to work out that the lim n->infinity (2n+2/3)/(3n+1) is 2/3. We were shown rules for limits as n tends to infinity but I don't think I've come across the above situation before.

Ok this is your function:

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac{2n-1}{3n+1}$$

Now, there's a shortcut. I don't know if you know it already but, if the powers are the same on the numerator and the denominator, just use the coefficients. Look at the 2n and the 3n. They have the same powers of 1 so use their coefficients 2 and 3. That will basically equal to 2/3. Hope that helps. Remember only when the powers are the same.

What you can also do is divide everything by the highest power in the denominator which is going to be n.

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac {{\frac{2n}{n}}-{\frac{1}{n}}}{\frac{3n}{n}+{\frac{1}{n}}}$$

Cancel and you're left with:

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac{{2}-\frac{1}{n}}{{3}+\frac{1}{n}}$$

Now plug zero into the n's and you're left with:

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac{2}{3}$$

The method you were taught seems complicated to me.

 

[shan]

Cool, that makes it much clearer and simpler. Thanks very much :)

 

[Ouabache]

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac{{2}-\frac{1}{n}}{{3}+\frac{1}{n}}$$

Now plug zero into the n's and you're left with:

$$\lim_{\substack{n\rightarrow \infty}}f(n)=\frac{2}{3}$$

I agree with jzq but with a small edition:
Since n approaches infinity in this question, I think jzq meant to say:

"Now plug zero in for (1/n)'s and you're left with:" ...

(that is because 1/infinity = 0).

A good online reference for "limit questions" can be viewed at:
http://www.jtaylor1142001.net/calcjat/Contents/CLimits.html

 

[Severian596]

Man...limits are fun, aren't they?

 

[shan]

A good online reference for "limit questions" can be viewed at:
http://www.jtaylor1142001.net/calcjat/Contents/CLimits.html

Thanks so much for that site, it's a godsend! :!)