Finding $limlimits_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)$
up vote
3
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What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.
calculus limits
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
add a comment |
up vote
3
down vote
favorite
What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.
calculus limits
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.
calculus limits
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
What is$$lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)?$$So it is$$lim_{n→∞}frac{n^3(sqrt{n^2+sqrt{n^4+1}})^2-(nsqrt{2})^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}.$$
I do not know what to do next, because my resuts is $∞$ but the answer from book is $dfrac{1}{4sqrt{2}}$.
calculus limits
calculus limits
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
edited Dec 4 at 10:18
Jyrki Lahtonen
107k12166364
107k12166364
asked Dec 4 at 9:59
Estera Gmiterek
306
asked Dec 4 at 9:59
Estera Gmiterek
306
asked Dec 4 at 9:59
Estera Gmiterek
306
306
try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39
add a comment |
try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39
try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39
add a comment |
5 Answers
5
active
oldest
votes
up vote
4
down vote
accepted
HINT
You only need to apply the trick twice, indeed we have that
$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$
and
$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
answered Dec 4 at 10:05
gimusi
91.9k84495
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
add a comment |
up vote
2
down vote
Let $1/n=h$
$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
add a comment |
up vote
1
down vote
The expedite way:
$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$
and the limit is
$$frac{sqrt2}8.$$
answered Dec 4 at 10:31
Yves Daoust
123k668219
add a comment |
up vote
0
down vote
First replace $1/x=h$ to find
$$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$
$$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$
$$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
add a comment |
up vote
0
down vote
Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
$$
lim_{tto0^+}frac{1}{t^3}left(
sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
right)=
lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.
Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$
answered Dec 4 at 10:28
egreg
176k1384198
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
HINT
You only need to apply the trick twice, indeed we have that
$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$
and
$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
answered Dec 4 at 10:05
gimusi
91.9k84495
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
add a comment |
up vote
4
down vote
accepted
HINT
You only need to apply the trick twice, indeed we have that
$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$
and
$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
answered Dec 4 at 10:05
gimusi
91.9k84495
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
HINT
You only need to apply the trick twice, indeed we have that
$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$
and
$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
answered Dec 4 at 10:05
gimusi
91.9k84495
HINT
You only need to apply the trick twice, indeed we have that
$$sqrt{n^2+sqrt{n^4+1}}-nsqrt{2}=(sqrt{n^2+sqrt{n^4+1}}-nsqrt{2})cdotfrac{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=$$$$=frac{n^2+sqrt{n^4+1}-2n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$
and
$$frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}=frac{sqrt{n^4+1}-n^2}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}cdot frac{sqrt{n^4+1}+n^2}{sqrt{n^4+1}+n^2}=$$$$=frac{1}{(sqrt{n^2+sqrt{n^4+1}}+nsqrt{2})(sqrt{n^4+1}+n^2)}$$
Can you conclude form here?
answered Dec 4 at 10:05
gimusi
91.9k84495
edited Dec 4 at 10:10
answered Dec 4 at 10:05
gimusi
91.9k84495
answered Dec 4 at 10:05
gimusi
91.9k84495
answered Dec 4 at 10:05
gimusi
91.9k84495
91.9k84495
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
add a comment |
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
Yes, thanks i understand !
– Estera Gmiterek
Dec 4 at 19:13
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
That’s nice! Well done, you are welcome! Bye
– gimusi
Dec 4 at 19:17
add a comment |
up vote
2
down vote
Let $1/n=h$
$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
add a comment |
up vote
2
down vote
Let $1/n=h$
$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $1/n=h$
$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
Let $1/n=h$
$$lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
$$=lim_{hto0^+}dfrac{1+sqrt{1+h^4}-2}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=lim_{hto0^+}dfrac{1+h^4-1}{h^4}cdotlim_{hto0^+}dfrac1{sqrt{1+h^4}+1}cdotlim_{hto0^+}dfrac1{sqrt{1+sqrt{1+h^4}}+sqrt2}$$
$$=dfrac1{(sqrt1+1)(sqrt{1+sqrt1}+sqrt2)}$$
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
edited Dec 4 at 10:33
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
answered Dec 4 at 10:08
lab bhattacharjee
222k15155273
222k15155273
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
add a comment |
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Nice way to simplify the manipulation. (+1)
– gimusi
Dec 4 at 10:12
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
Sorry, but the final result is wrong
– egreg
Dec 4 at 10:28
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
@egreg, Thanks for your feedback. Rectfied.
– lab bhattacharjee
Dec 4 at 10:33
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
Related : math.stackexchange.com/questions/524288/…
– lab bhattacharjee
Dec 7 at 5:10
add a comment |
up vote
1
down vote
The expedite way:
$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$
and the limit is
$$frac{sqrt2}8.$$
answered Dec 4 at 10:31
Yves Daoust
123k668219
add a comment |
up vote
1
down vote
The expedite way:
$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$
and the limit is
$$frac{sqrt2}8.$$
answered Dec 4 at 10:31
Yves Daoust
123k668219
add a comment |
up vote
1
down vote
up vote
1
down vote
The expedite way:
$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$
and the limit is
$$frac{sqrt2}8.$$
answered Dec 4 at 10:31
Yves Daoust
123k668219
The expedite way:
$$sqrt{1+sqrt{1+n^{-4}}}=sqrt{1+1+dfrac12n^{-4}+o(n^{-4})}=sqrt2sqrt{1+dfrac14n^{-4}+o(n^{-4})}=sqrt2left(1+dfrac18n^{-4}+o(n^{-4})right)$$
and the limit is
$$frac{sqrt2}8.$$
answered Dec 4 at 10:31
Yves Daoust
123k668219
answered Dec 4 at 10:31
Yves Daoust
123k668219
answered Dec 4 at 10:31
Yves Daoust
123k668219
answered Dec 4 at 10:31
Yves Daoust
123k668219
123k668219
add a comment |
add a comment |
up vote
0
down vote
First replace $1/x=h$ to find
$$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$
$$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$
$$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
add a comment |
up vote
0
down vote
First replace $1/x=h$ to find
$$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$
$$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$
$$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
add a comment |
up vote
0
down vote
up vote
0
down vote
First replace $1/x=h$ to find
$$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$
$$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$
$$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
First replace $1/x=h$ to find
$$L=lim_{n→∞}n^3(sqrt{n^2+sqrt{n^4+1}}-nsqrt2)=lim_{hto0^+}dfrac{sqrt{1+sqrt{1+h^4}}-sqrt2}{h^4}$$
Let $sqrt{1+sqrt{1+h^4}}=y+sqrt2implies1+sqrt{1+h^4}=(sqrt2+y)^2=2+2sqrt2y+y^2$
$$1+h^4=(1+y(2sqrt2+y))^2=1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2$$
$$L=lim_{yto0}dfrac{1+2y(2sqrt2+y)+y^2(2sqrt2+y)^2-1}y=?$$
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
answered Dec 4 at 10:16
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
up vote
0
down vote
Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
$$
lim_{tto0^+}frac{1}{t^3}left(
sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
right)=
lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.
Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$
answered Dec 4 at 10:28
egreg
176k1384198
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
add a comment |
up vote
0
down vote
Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
$$
lim_{tto0^+}frac{1}{t^3}left(
sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
right)=
lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.
Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$
answered Dec 4 at 10:28
egreg
176k1384198
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
add a comment |
up vote
0
down vote
up vote
0
down vote
Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
$$
lim_{tto0^+}frac{1}{t^3}left(
sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
right)=
lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.
Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$
answered Dec 4 at 10:28
egreg
176k1384198
Formally substitute $n=1/t$; if the function you get has a limit for $tto0^+$, then it is the same as the limit you are looking for. So consider
$$
lim_{tto0^+}frac{1}{t^3}left(
sqrt{frac{1}{t^2}+sqrt{frac{1}{t^4}+1}}-frac{sqrt{2}}{t}
right)=
lim_{tto0^+}frac{sqrt{1+sqrt{1+t^4}}-sqrt{2}}{t^4}
$$
Now the dependency is only on $t^4$, so the limit is the same as
$$
lim_{uto0^+}frac{sqrt{1+sqrt{1+u}}-sqrt{2}}{u}
$$
which is the derivative at $0$ of $f(u)=sqrt{1+sqrt{1+u}}$.
Since $$f'(u)=frac{1}{2sqrt{1+sqrt{1+u}}}frac{1}{2sqrt{1+u}}$$ we have $$f'(0)=frac{1}{4sqrt{2}}$$
answered Dec 4 at 10:28
egreg
176k1384198
edited Dec 4 at 12:36
answered Dec 4 at 10:28
egreg
176k1384198
answered Dec 4 at 10:28
egreg
176k1384198
answered Dec 4 at 10:28
egreg
176k1384198
176k1384198
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
add a comment |
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
As always I don't agree that you give a full answer. In my opinion it is not useful for the asker nor for the site. I don't downvote because it is of course a correct and good answer. But virtually I give you a "-1".
– gimusi
Dec 4 at 10:31
add a comment |
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try putting n=1/x.then apply binomial expansions.
– maveric
Dec 4 at 10:07
@maveric That's of course an effective method but we don't really need that on this case.
– gimusi
Dec 4 at 10:08
How can be your result $infty$ from here $$lim_{n→∞}frac{n^3(n^2+sqrt{n^4+1}-2n^2)}{sqrt{n^2+sqrt{n^4+1}}+nsqrt{2}}$$ Note that at the numerator we have an indeterminate form $infty-infty$ therefore you can't conclude form here.
– gimusi
Dec 4 at 10:14
@gimusi. Using the binomial expansion is a good way if you want to know how is approached the limit (this is just building the Taylor expansion). I hope and wish that we could have a discussion about this kind of problems one of these days. Cheers.
– Claude Leibovici
Dec 4 at 10:37
@ClaudeLeibovici Yes of course, I also like Taylor's very much but in that case I think that it suffices operate by agebraic methods also I think thet the OP is looking for such kind of solution. Anyway the suggestion is good as an alternative approach! Cheers
– gimusi
Dec 4 at 10:39