1. ## Limits of sequences

What are the limits of the follwing sequences as n tends towards infinity?

(i) $\sqrt {{n}^{2}+1000}-n$

(ii) $\left( \left( n! \right) ^{{n}^{-2}} \right) ^{-1}$

(iii) ${\frac {{n}^{100}}{{ 1.001}^{n}}}$

(iv) ${\frac {n \left( 2\,n+3 \right) \left( 3\,n+4 \right) +2}{4\, \left(
n \left( 6\,n+7 \right) \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}$

Im not really sure how to do the others, they are part of a past exam paper and there is no mark scheme, so when i get to an answer im not sure if it is right or not!

2. I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?

3. Originally Posted by Mathman87
I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?
(i) is 0

4. (ii) re-write the equation $x^{-1}=\frac{1}{x}$

5. (iv) $\frac{6n^3}{144n^4}=\frac{1}{24n}$
So the limit is 0

6. Originally Posted by Mathman87
I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?

Using intuition is a first step to solving problems , then you may solve them by following your intuition .

For example , for question (i) , intuitively it is zero , we can do something to prove it zero . We know $\frac{1}{\infty}$ is zero so can we do something to rewrite the limit in this form ?

$\sqrt{n^2 + 1000} - n$

$= (\sqrt{n^2 + 1000} - n ) \cdot \frac{\sqrt{n^2 + 1000} + n }{\sqrt{n^2 + 1000} + n }$

$= \frac{(n^2+1000)-n^2 }{ \sqrt{n^2 + 1000} + n }$ .

7. (iii) is 0 believe it or not.

$\frac {n^{100}}{1.001^n} = \Big(\frac {n^{\frac{100}{n}}}{1.001}\Big)^n$

$\frac{100}{n}\to 0 \implies \Big(\frac {n^{\frac{100}{n}}}{1.001}\Big)^n \to 0.$

8. Originally Posted by Anonymous1
(iii) is 0 believe it or not.
I believe it! Maple told me so.

10. Originally Posted by Mathman87
What are the limits of the follwing sequences as n tends towards infinity?

(i) $\sqrt {{n}^{2}+1000}-n$
Multiply numerator and denominator by $\sqrt{n^2+ 1000}+ n$:
$(\sqrt{n^2+ 1000}- n)\frac{\sqrt{n^2+ 1000}+ n}{\sqrt{n^2+ 1000}+ n}= \frac{n^2+ 1000- n^2}{\sqrt{n^2+ 1000}+ n}= \frac{1000}{\sqrt{n^2+ 1000}+ n}$.

(ii) $\left( \left( n! \right) ^{{n}^{-2}} \right) ^{-1}$
This is the same as $\frac{1}{(n!)^{n^2}}$. The numerator is 1 while the denominator gets bigger and bigger without bound.

(iii) ${\frac {{n}^{100}}{{ 1.001}^{n}}}$
Probably the simplest thing to do is use L'Hopital's rule- 100 times!

(iv) ${\frac {n \left( 2\,n+3 \right) \left( 3\,n+4 \right) +2}{4\, \left(
n \left( 6\,n+7 \right) \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}$

If you were to multiply out the numerator, the leading term would be 6n^3. Doing the same in the denominator the leading term would be 144n^4. Divide both numerator and denominator by $n^4$.