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limits

hi, i dont understand how to do this limit, for n n \to \infty


d) (2n+1n+1)2n (\frac{2n+1}{n+1})^{2n}
(edited 5 years ago)
plz help its 3:39 am and im not going to sleep until i do this
Original post by KloppOClock
hi, i dont understand how to do this limit, for n n \to \infty


d) (2n+1n+1)2n (\frac{2n+1}{n+1})^{2n}


divide numerator and denominator by n, to get (2+1n1+1n)2n ( \frac{2+ \frac{1}{n}}{1+ \frac{1}{n}} )^{2n} clearly inside the bracket goes to 2, power increases thus limit doesn't exist (tends to infinity)
(edited 5 years ago)
try to use the fact that

(2n+1)/(n+1)=(n+(n+1))/(n+1)=1+n/(n+1) , which may be more familiar to you
just tried that and it didnt help me :frown:

the answer sheet looks like its doign squeeze theorem but its so vague i cant follow it
(edited 5 years ago)
Additionally, this rule applies for exponents

for x approaching some number/infinity
lim(ax) = alim(x)
Original post by TheTroll73
try to use the fact that

(2n+1)/(n+1)=(n+(n+1))/(n+1)=1+n/(n+1) , which may be more familiar to you


okay so that gets me to;

(1+nn+1)2n (1+\frac{n}{n+1})^2n which sorta looks like a home bargains version of 'e' .

can i just evaluate it there to get (1+1)^infinity and say it has no limit (which it doesnt) ?
Original post by zetamcfc
divide numerator and denominator by n, to get (2+1n1+1n)2n ( \frac{2+ \frac{1}{n}}{1+ \frac{1}{n}} )^{2n} clearly inside the bracket goes to 2, power increases thus limit doesn't exist (tends to infinity)


ah okay thats one good easier way of doing it, could you help me understand where the mark scheme did it?

they proved 2n+1n+1 \frac{2n+1}{n+1} was bigger than 3/2 for all n > 0 some how then said (3/2)^2n = (9/4)^n has no limit since 9/4 bigger than 1

i dont get this
Original post by KloppOClock
just tried that and it didnt help me :frown:

the answer sheet looks like its doign squeeze theorem but its so vague i cant follow it


Could you post a picture of the answer sheet.
Original post by KloppOClock
okay so that gets me to;

(1+nn+1)2n (1+\frac{n}{n+1})^2n which sorta looks like a home bargains version of 'e' .

can i just evaluate it there to get (1+1)^infinity and say it has no limit (which it doesnt) ?


yes since as n approaches infinity,

lim(1+n/(n+1))^2n = 2^lim(2n) which yields no limit
Original post by KloppOClock
ah okay thats one good easier way of doing it, could you help me understand where the mark scheme did it?

they proved 2n+1n+1 \frac{2n+1}{n+1} was bigger than 3/2 for all n > 0 some how then said (3/2)^2n = (9/4)^n has no limit since 9/4 bigger than 1

i dont get this


So what they do is that because, for all n>0 , a_n=(3/2)^(2n) < b_n=((2n+1)/(n+1))^(2n) this means that lim n to infinity of a_n < lim n to infinity of b_n which means that as lim a_n goes to infinity, that also lim b_n must also go to infinity.
(edited 5 years ago)
Untitled.jpg

Original post by zetamcfc
Could you post a picture of the answer sheet.
Original post by zetamcfc
So what they do is that because, for all n>0 , a_n=(3/2)^(2n) < b_n=((2n+1)/(n+1))^(2n) this means that lim n to infinity of a_n < lim n to infinity of b_n which means that as lim a_n goes to infinity, that also lim b_n must also go to infinity.


i see now, dem cheeky feckers
That could be a 1 to the power of infinity limit. The easiest way of solving those is by using the e to the power of the limit minus one formula:
IMG_20180717_052015.jpg
(edited 5 years ago)
Original post by Stakanovicius
Thats a 1 to the power of infinity type of limit. The easiest way of solving it is by using the e to the power of the limit minus one formula:
IMG_20180717_052015.jpg


that would seem easy but i dont think weve been taught that so ill probably ignore its existence for now, thanks tho
I have been forced to use another much time consuming method throughout most of my last two years of highschool, but at the end of my last year I found out about this formula and thought about the amount of time I had spent for nothing xdddd
Original post by Stakanovicius
That could be a 1 to the power of infinity limit. How? limn2n+1n\lim_{n \to \infty} \dfrac{2n+1}{n} is 2, not 1.

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