Limits of sequences

Apr 2010
30
0
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) \(\displaystyle {\frac {n \left( 2\,n+3 \right) \left( 3\,n+4 \right) +2}{4\, \left(
n \left( 6\,n+7 \right) \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}\)

I had a go at (iv) and got an answer 1/4....

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!
 
Apr 2010
30
0
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

MHF Hall of Honor
Mar 2010
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582
Florida
(ii) re-write the equation \(\displaystyle x^{-1}=\frac{1}{x}\)
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
(iv)\(\displaystyle \frac{6n^3}{144n^4}=\frac{1}{24n}\)
So the limit is 0
 
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simplependulum

MHF Hall of Honor
Jan 2009
715
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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 ?

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

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

\(\displaystyle = \frac{(n^2+1000)-n^2 }{ \sqrt{n^2 + 1000} + n }\) .
 
Nov 2009
517
130
Big Red, NY
(iii) is 0 believe it or not.

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

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

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Misread question
 
Last edited:

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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) \(\displaystyle {\frac {n \left( 2\,n+3 \right) \left( 3\,n+4 \right) +2}{4\, \left(
n \left( 6\,n+7 \right) \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}\)

I had a go at (iv) and got an answer 1/4....
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 \(\displaystyle n^4\).

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!