Limit as n approaches infinity of (3n^2 + n + 1)/(5n^3 -2n + 2)

[Lo.Lee.Ta.]

1. lim (3n² + n + 1)/(5n³ - 2n + 2)
n→∞

2. In order to solve this problem, do you just think about what happens when n is replaced with a really big number?

So, in this case, the numerator only has an n² and an n, but the denominator has an n³ and an n...

So the bottom would always be bigger than the top. So the limit goes to zero?

Thanks.

 

[Ray Vickson]

1. lim (3n² + n + 1)/(5n³ - 2n + 2)
n→∞

2. In order to solve this problem, do you just think about what happens when n is replaced with a really big number?

So, in this case, the numerator only has an n² and an n, but the denominator has an n³ and an n...

So the bottom would always be bigger than the top. So the limit goes to zero?

Thanks.

The expression goes to zero; the limit does not "go to zero"; it IS zero. (Limits are what other things go to; they themselves do not go to anything.)

 

[Skins]

You could also apply L'Hopital's rule to this limit as in it's fits a valid indeterminate form

\frac{\infty}{\infty}

Of course it also depends on whether you learned L'Hopitals rule yet.

 

[ehild]

1. lim (3n² + n + 1)/(5n³ - 2n + 2)
n→∞

2. In order to solve this problem, do you just think about what happens when n is replaced with a really big number?

So, in this case, the numerator only has an n² and an n, but the denominator has an n³ and an n...

So the bottom would always be bigger than the top. So the limit goes to zero?

Thanks.

The standard method to find the limit of rational functions is to factor out the highest power of n both from numerator and denominator.

\frac{n^2(3+1/n+1/n^2)}{n^3(5-2/n+2/n^2)}
Simplify by n^2, and take the limit separately for each terms. You are left with the limit of 3/(5n) .

ehild

 

[Lo.Lee.Ta.]

Hopefully, I'm not making some stupid mistake, but I get this using your method, ehild:

n²(3 + 1/n + 1/n²)/(n³(5 - 2/n² + 2/n³))

= (3 + 1/n + 1/n²)/(n(5 - 2/n² + 2/n³)

= 3/5 - 1/2 + 1/2 = 3/5

Did I make some mistakes, or is this what the expression goes to...?

Thanks! :)

 

[ehild]

Hopefully, I'm not making some stupid mistake, but I get this using your method, ehild:

n²(3 + 1/n + 1/n²)/(n³(5 - 2/n² + 2/n³))

= (3 + 1/n + 1/n²)/(n(5 - 2/n² + 2/n³))

= [STRIKE]3/5 - 1/2 + 1/2 = 3/5[/STRIKE]

Did I make some mistakes, or is this what the expression goes to...?

Thanks! :)

You omitted the "n" from the denominator and left out a ")". And the limits of all 1/n, 1/n^2, 1/n^3 are zero when n goes to infinity.

ehild

 

[Lo.Lee.Ta.]

Oh, yeah! I did omit that n in the denominator! :/

Now I get the same thing as you: 3/5n

and that goes to 0 when n→∞

Thanks! :D

 

[Mark44]

Hopefully, I'm not making some stupid mistake, but I get this using your method, ehild:

n²(3 + 1/n + 1/n²)/(n³(5 - 2/n² + 2/n³))

= (3 + 1/n + 1/n²)/(n(5 - 2/n² + 2/n³)

= 3/5 - 1/2 + 1/2 = 3/5

Did I make some mistakes, or is this what the expression goes to...?

Thanks! :)

You omitted the "n" from the denominator and left out a ")". And the limits of all 1/n, 1/n^2, 1/n^3 are zero when n goes to infinity.

In addition to what ehild said, you are making a silly algebra error. The following is NOT TRUE[/color].
$$ \frac{a + b}{c + d} = \frac{a}{c} + \frac{b}{d}$$

For example, (1 + 1)/(2 + 2) ≠ 1/2 + 1/2 = 1

 

[Lo.Lee.Ta.]

Okay, right! ...I was sort of skeptical when I wrote that out the first time... :/

(3 + 1/n + 1/n²)/(5n - 2/n + 2/n²)

So at this point, should I analyze every term separately?

I mean, look at 1/n, 1/n², 2/n, and 2/n², and know that they all go to zero- so replace them with zero.

That would mean I'm left with 3/5n, and that goes to 0.

Is that the way I should think of it?

Thanks! :)

 

[ehild]

Yes, analyse every term separately. You know that the limit of a sum/difference is the sum /difference of limits, multiplying with a constant multiplies the limit, and the limit of a fraction is equal to the limit of the nominator divided by the limit of the denominator (except when it is zero).

ehild