# Limits of sequences

#### Mathman87

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

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

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

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

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

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!

#### 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?

#### dwsmith

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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

#### dwsmith

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

#### dwsmith

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

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#### simplependulum

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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 $$\displaystyle \frac{1}{\infty}$$ is zero so can we do something to rewrite the limit in this form ?

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

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

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

#### Anonymous1

(iii) is 0 believe it or not.

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

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

#### dwsmith

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MHF Helper

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#### HallsofIvy

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

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

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

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

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

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 $$\displaystyle n^4$$.