Mixing
Which function grows faster?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Which function grows faster
$ð‘“(ð‘›)= 2^ð‘›^2+3ð‘›$ and $ð‘”(ð‘›) = 2^ð‘›+1$
by using the limit theorem I will first simplify
then I will just get $$lim_n to infty dfrac2^n^2+3n2^n+1=lim_n to infty 2^n^2+3n-n-1=lim_n to infty 2^n^2+2n-1=infty$$
Is this enough?
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!
limits logarithms asymptotics
asked 52 mins ago
NANA
327
add a comment |Â
up vote
2
down vote
favorite
Which function grows faster
$ð‘“(ð‘›)= 2^ð‘›^2+3ð‘›$ and $ð‘”(ð‘›) = 2^ð‘›+1$
by using the limit theorem I will first simplify
then I will just get $$lim_n to infty dfrac2^n^2+3n2^n+1=lim_n to infty 2^n^2+3n-n-1=lim_n to infty 2^n^2+2n-1=infty$$
Is this enough?
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!
limits logarithms asymptotics
asked 52 mins ago
NANA
327
I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
1
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Which function grows faster
$ð‘“(ð‘›)= 2^ð‘›^2+3ð‘›$ and $ð‘”(ð‘›) = 2^ð‘›+1$
by using the limit theorem I will first simplify
then I will just get $$lim_n to infty dfrac2^n^2+3n2^n+1=lim_n to infty 2^n^2+3n-n-1=lim_n to infty 2^n^2+2n-1=infty$$
Is this enough?
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!
limits logarithms asymptotics
asked 52 mins ago
NANA
327
Which function grows faster
$ð‘“(ð‘›)= 2^ð‘›^2+3ð‘›$ and $ð‘”(ð‘›) = 2^ð‘›+1$
by using the limit theorem I will first simplify
then I will just get $$lim_n to infty dfrac2^n^2+3n2^n+1=lim_n to infty 2^n^2+3n-n-1=lim_n to infty 2^n^2+2n-1=infty$$
Is this enough?
I say it will go then to infinity so the $f(n)$ is growing faster? I am asking this question because I have to find it by using limit but I didn't need to use l'hopital rule!
limits logarithms asymptotics
limits logarithms asymptotics
asked 52 mins ago
NANA
327
asked 52 mins ago
NANA
327
edited 42 mins ago
asked 52 mins ago
NANA
327
asked 52 mins ago
NANA
327
asked 52 mins ago
NANA
327
327
I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
1
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago
add a comment |Â
I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
1
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago
I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
1
1
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$lim_n to infty frac2^n^2+3n2^n+1$$
It is enough to divide both the numerator and denominator by $2^n$.
$$lim_n to infty fracfrac2^n^2+3n2^nfrac2^n+12^n = lim_n to infty frac2^n^2+3n-n2^n-n+frac12^n = lim_n to infty frac2^n^2+2n1+frac12^n$$
As $n to infty$, it becomes clear that the limit tends to $infty$ since the numerator tends to $infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
answered 44 mins ago
KM101
969110
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
add a comment |Â
up vote
2
down vote
It is $$frac2^n^2cdot 2^3n2^nleft(1+frac12^nright)=frac2^n^2+2n1+frac12^n$$
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
add a comment |Â
up vote
2
down vote
HINT
You conclusion is correct but that step is wrong
$$lim_n to infty= dfrac2^n^2+3n2^n+1colorred=lim_n to infty 2^n^2+3n-n-1$$
you could use that $2^n+1le 2^n+1$ and therefore
$$dfrac2^n^2+3n2^n+1ge dfrac2^n^2+3n2^n+1$$
Update after editing
For $g(n)=2^n+1$ your method is absolutely correct.
What about $f(n)=2^n^2+3n$ and $g(n)=3^n+1$?
answered 50 mins ago
gimusi
79.3k73990
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$lim_n to infty frac2^n^2+3n2^n+1$$
It is enough to divide both the numerator and denominator by $2^n$.
$$lim_n to infty fracfrac2^n^2+3n2^nfrac2^n+12^n = lim_n to infty frac2^n^2+3n-n2^n-n+frac12^n = lim_n to infty frac2^n^2+2n1+frac12^n$$
As $n to infty$, it becomes clear that the limit tends to $infty$ since the numerator tends to $infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
answered 44 mins ago
KM101
969110
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
add a comment |Â
up vote
3
down vote
accepted
Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$lim_n to infty frac2^n^2+3n2^n+1$$
It is enough to divide both the numerator and denominator by $2^n$.
$$lim_n to infty fracfrac2^n^2+3n2^nfrac2^n+12^n = lim_n to infty frac2^n^2+3n-n2^n-n+frac12^n = lim_n to infty frac2^n^2+2n1+frac12^n$$
As $n to infty$, it becomes clear that the limit tends to $infty$ since the numerator tends to $infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
answered 44 mins ago
KM101
969110
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$lim_n to infty frac2^n^2+3n2^n+1$$
It is enough to divide both the numerator and denominator by $2^n$.
$$lim_n to infty fracfrac2^n^2+3n2^nfrac2^n+12^n = lim_n to infty frac2^n^2+3n-n2^n-n+frac12^n = lim_n to infty frac2^n^2+2n1+frac12^n$$
As $n to infty$, it becomes clear that the limit tends to $infty$ since the numerator tends to $infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
answered 44 mins ago
KM101
969110
Before Edit:
Your idea was correct, but you didn’t simplify the limit properly.
$$lim_n to infty frac2^n^2+3n2^n+1$$
It is enough to divide both the numerator and denominator by $2^n$.
$$lim_n to infty fracfrac2^n^2+3n2^nfrac2^n+12^n = lim_n to infty frac2^n^2+3n-n2^n-n+frac12^n = lim_n to infty frac2^n^2+2n1+frac12^n$$
As $n to infty$, it becomes clear that the limit tends to $infty$ since the numerator tends to $infty$ while the denominator tends to $1$.
After Edit: Yes, your way is correct.
answered 44 mins ago
KM101
969110
edited 18 mins ago
answered 44 mins ago
KM101
969110
answered 44 mins ago
KM101
969110
answered 44 mins ago
KM101
969110
969110
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
add a comment |Â
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
I forget to add brackets on the second function, it must be $2^n+1$ @KM101
– NANA
42 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
Well, I guess all the answers are pointless now. :-)
– KM101
33 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
sorry for that! and thanks, then shall I put the first question too so that it wouldn't be pointless !! :) @KM101
– NANA
32 mins ago
add a comment |Â
up vote
2
down vote
It is $$frac2^n^2cdot 2^3n2^nleft(1+frac12^nright)=frac2^n^2+2n1+frac12^n$$
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
add a comment |Â
up vote
2
down vote
It is $$frac2^n^2cdot 2^3n2^nleft(1+frac12^nright)=frac2^n^2+2n1+frac12^n$$
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is $$frac2^n^2cdot 2^3n2^nleft(1+frac12^nright)=frac2^n^2+2n1+frac12^n$$
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
It is $$frac2^n^2cdot 2^3n2^nleft(1+frac12^nright)=frac2^n^2+2n1+frac12^n$$
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
answered 48 mins ago
Dr. Sonnhard Graubner
71.2k32863
71.2k32863
add a comment |Â
add a comment |Â
up vote
2
down vote
HINT
You conclusion is correct but that step is wrong
$$lim_n to infty= dfrac2^n^2+3n2^n+1colorred=lim_n to infty 2^n^2+3n-n-1$$
you could use that $2^n+1le 2^n+1$ and therefore
$$dfrac2^n^2+3n2^n+1ge dfrac2^n^2+3n2^n+1$$
Update after editing
For $g(n)=2^n+1$ your method is absolutely correct.
What about $f(n)=2^n^2+3n$ and $g(n)=3^n+1$?
answered 50 mins ago
gimusi
79.3k73990
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
add a comment |Â
up vote
2
down vote
HINT
You conclusion is correct but that step is wrong
$$lim_n to infty= dfrac2^n^2+3n2^n+1colorred=lim_n to infty 2^n^2+3n-n-1$$
you could use that $2^n+1le 2^n+1$ and therefore
$$dfrac2^n^2+3n2^n+1ge dfrac2^n^2+3n2^n+1$$
Update after editing
For $g(n)=2^n+1$ your method is absolutely correct.
What about $f(n)=2^n^2+3n$ and $g(n)=3^n+1$?
answered 50 mins ago
gimusi
79.3k73990
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT
You conclusion is correct but that step is wrong
$$lim_n to infty= dfrac2^n^2+3n2^n+1colorred=lim_n to infty 2^n^2+3n-n-1$$
you could use that $2^n+1le 2^n+1$ and therefore
$$dfrac2^n^2+3n2^n+1ge dfrac2^n^2+3n2^n+1$$
Update after editing
For $g(n)=2^n+1$ your method is absolutely correct.
What about $f(n)=2^n^2+3n$ and $g(n)=3^n+1$?
answered 50 mins ago
gimusi
79.3k73990
HINT
You conclusion is correct but that step is wrong
$$lim_n to infty= dfrac2^n^2+3n2^n+1colorred=lim_n to infty 2^n^2+3n-n-1$$
you could use that $2^n+1le 2^n+1$ and therefore
$$dfrac2^n^2+3n2^n+1ge dfrac2^n^2+3n2^n+1$$
Update after editing
For $g(n)=2^n+1$ your method is absolutely correct.
What about $f(n)=2^n^2+3n$ and $g(n)=3^n+1$?
answered 50 mins ago
gimusi
79.3k73990
edited 31 mins ago
answered 50 mins ago
gimusi
79.3k73990
answered 50 mins ago
gimusi
79.3k73990
answered 50 mins ago
gimusi
79.3k73990
79.3k73990
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
add a comment |Â
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
I have this question too! but couldn't find any idea how to start solving it, because of the 3 ! any hint ?? @gimusi
– NANA
21 mins ago
1
1
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
@NANA We could use $3^n+1le 4^n+1$.
– gimusi
4 mins ago
add a comment |Â
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I reformatted your limits, but I didn't change the arithmetic (or at least, I didn't intend to). You appear to have mishandled the exponents.
– lulu
48 mins ago
@lulu could you see the question again I edit the 1 , must be on the power of 2
– NANA
41 mins ago
1
Yes. With that edit , your arithmetic is correct. And your argument is sufficient.
– lulu
31 mins ago