Mixing
Limit with binomials
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$$lim_xtoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n.
$$
I tried to do the binomial expansion and then operate only with the odds but I couldn't find the limit.Can somebody help me,please?
calculus limits
asked 2 hours ago
Andrei Gabor
435
add a comment |Â
up vote
2
down vote
favorite
$$lim_xtoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n.
$$
I tried to do the binomial expansion and then operate only with the odds but I couldn't find the limit.Can somebody help me,please?
calculus limits
asked 2 hours ago
Andrei Gabor
435
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$lim_xtoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n.
$$
I tried to do the binomial expansion and then operate only with the odds but I couldn't find the limit.Can somebody help me,please?
calculus limits
asked 2 hours ago
Andrei Gabor
435
$$lim_xtoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n.
$$
I tried to do the binomial expansion and then operate only with the odds but I couldn't find the limit.Can somebody help me,please?
calculus limits
calculus limits
asked 2 hours ago
Andrei Gabor
435
asked 2 hours ago
Andrei Gabor
435
edited 34 mins ago
asked 2 hours ago
Andrei Gabor
435
asked 2 hours ago
Andrei Gabor
435
asked 2 hours ago
Andrei Gabor
435
435
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago
add a comment |Â
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
You have
$$frac(2x+sqrtx^2-1)^n + (2x-sqrtx^2-1)^nx^n= left(2 + sqrt1-frac1x^2right)^n +left(2 - sqrt1-frac1x^2right)^n geq left(2 + sqrt1-frac1x^2right)^n$$
which tends to $+infty$ when $n rightarrow +infty$.
answered 2 hours ago
TheSilverDoe
1,81411
add a comment |Â
up vote
2
down vote
Note that
$$frac(2xpmsqrtx^2-1)^nx^n=left(2pmsqrt1-frac1x^2right)^n $$
So at least one of the summands is $ge 2^n$, and the other still positive.
Even if we allow complex $x$, we have an expression of the form $a^n+b^n$ with $a+b=4$, hence either one of $|a|,|b|$ is larger than the other and then larger then $2$ (thus leading to divergence), or $|a|=|b|$ and then the real parts of $a^n$ and $b^n$ are always equal, but are unbounded.
answered 2 hours ago
Hagen von Eitzen
269k21261485
X goes to infinity...sory
– Andrei Gabor
33 mins ago
add a comment |Â
up vote
1
down vote
$$lim_ntoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n$$
$$= lim_ntoinftybigg (2+sqrt1- 1/x^2bigg)^n+lim_ntoinftybigg(2-sqrt1- 1/x^2bigg )^n = infty$$
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
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
You have
$$frac(2x+sqrtx^2-1)^n + (2x-sqrtx^2-1)^nx^n= left(2 + sqrt1-frac1x^2right)^n +left(2 - sqrt1-frac1x^2right)^n geq left(2 + sqrt1-frac1x^2right)^n$$
which tends to $+infty$ when $n rightarrow +infty$.
answered 2 hours ago
TheSilverDoe
1,81411
add a comment |Â
up vote
3
down vote
You have
$$frac(2x+sqrtx^2-1)^n + (2x-sqrtx^2-1)^nx^n= left(2 + sqrt1-frac1x^2right)^n +left(2 - sqrt1-frac1x^2right)^n geq left(2 + sqrt1-frac1x^2right)^n$$
which tends to $+infty$ when $n rightarrow +infty$.
answered 2 hours ago
TheSilverDoe
1,81411
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You have
$$frac(2x+sqrtx^2-1)^n + (2x-sqrtx^2-1)^nx^n= left(2 + sqrt1-frac1x^2right)^n +left(2 - sqrt1-frac1x^2right)^n geq left(2 + sqrt1-frac1x^2right)^n$$
which tends to $+infty$ when $n rightarrow +infty$.
answered 2 hours ago
TheSilverDoe
1,81411
You have
$$frac(2x+sqrtx^2-1)^n + (2x-sqrtx^2-1)^nx^n= left(2 + sqrt1-frac1x^2right)^n +left(2 - sqrt1-frac1x^2right)^n geq left(2 + sqrt1-frac1x^2right)^n$$
which tends to $+infty$ when $n rightarrow +infty$.
answered 2 hours ago
TheSilverDoe
1,81411
answered 2 hours ago
TheSilverDoe
1,81411
answered 2 hours ago
TheSilverDoe
1,81411
answered 2 hours ago
TheSilverDoe
1,81411
1,81411
add a comment |Â
add a comment |Â
up vote
2
down vote
Note that
$$frac(2xpmsqrtx^2-1)^nx^n=left(2pmsqrt1-frac1x^2right)^n $$
So at least one of the summands is $ge 2^n$, and the other still positive.
Even if we allow complex $x$, we have an expression of the form $a^n+b^n$ with $a+b=4$, hence either one of $|a|,|b|$ is larger than the other and then larger then $2$ (thus leading to divergence), or $|a|=|b|$ and then the real parts of $a^n$ and $b^n$ are always equal, but are unbounded.
answered 2 hours ago
Hagen von Eitzen
269k21261485
X goes to infinity...sory
– Andrei Gabor
33 mins ago
add a comment |Â
up vote
2
down vote
Note that
$$frac(2xpmsqrtx^2-1)^nx^n=left(2pmsqrt1-frac1x^2right)^n $$
So at least one of the summands is $ge 2^n$, and the other still positive.
Even if we allow complex $x$, we have an expression of the form $a^n+b^n$ with $a+b=4$, hence either one of $|a|,|b|$ is larger than the other and then larger then $2$ (thus leading to divergence), or $|a|=|b|$ and then the real parts of $a^n$ and $b^n$ are always equal, but are unbounded.
answered 2 hours ago
Hagen von Eitzen
269k21261485
X goes to infinity...sory
– Andrei Gabor
33 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that
$$frac(2xpmsqrtx^2-1)^nx^n=left(2pmsqrt1-frac1x^2right)^n $$
So at least one of the summands is $ge 2^n$, and the other still positive.
Even if we allow complex $x$, we have an expression of the form $a^n+b^n$ with $a+b=4$, hence either one of $|a|,|b|$ is larger than the other and then larger then $2$ (thus leading to divergence), or $|a|=|b|$ and then the real parts of $a^n$ and $b^n$ are always equal, but are unbounded.
answered 2 hours ago
Hagen von Eitzen
269k21261485
Note that
$$frac(2xpmsqrtx^2-1)^nx^n=left(2pmsqrt1-frac1x^2right)^n $$
So at least one of the summands is $ge 2^n$, and the other still positive.
Even if we allow complex $x$, we have an expression of the form $a^n+b^n$ with $a+b=4$, hence either one of $|a|,|b|$ is larger than the other and then larger then $2$ (thus leading to divergence), or $|a|=|b|$ and then the real parts of $a^n$ and $b^n$ are always equal, but are unbounded.
answered 2 hours ago
Hagen von Eitzen
269k21261485
edited 1 hour ago
answered 2 hours ago
Hagen von Eitzen
269k21261485
answered 2 hours ago
Hagen von Eitzen
269k21261485
answered 2 hours ago
Hagen von Eitzen
269k21261485
269k21261485
X goes to infinity...sory
– Andrei Gabor
33 mins ago
add a comment |Â
X goes to infinity...sory
– Andrei Gabor
33 mins ago
X goes to infinity...sory
– Andrei Gabor
33 mins ago
X goes to infinity...sory
– Andrei Gabor
33 mins ago
add a comment |Â
up vote
1
down vote
$$lim_ntoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n$$
$$= lim_ntoinftybigg (2+sqrt1- 1/x^2bigg)^n+lim_ntoinftybigg(2-sqrt1- 1/x^2bigg )^n = infty$$
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
add a comment |Â
up vote
1
down vote
$$lim_ntoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n$$
$$= lim_ntoinftybigg (2+sqrt1- 1/x^2bigg)^n+lim_ntoinftybigg(2-sqrt1- 1/x^2bigg )^n = infty$$
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$lim_ntoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n$$
$$= lim_ntoinftybigg (2+sqrt1- 1/x^2bigg)^n+lim_ntoinftybigg(2-sqrt1- 1/x^2bigg )^n = infty$$
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
$$lim_ntoinfty frac(2x+sqrtx^2-1)^n+(2x-sqrtx^2-1)^nx^n$$
$$= lim_ntoinftybigg (2+sqrt1- 1/x^2bigg)^n+lim_ntoinftybigg(2-sqrt1- 1/x^2bigg )^n = infty$$
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
answered 1 hour ago
Mohammad Riazi-Kermani
33.3k41854
33.3k41854
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2940895%2flimit-with-binomials%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Why do you think you need the binomial expansion? Complex values of $x$?..
– metamorphy
2 hours ago
X goes to infinity...sorry
– Andrei Gabor
25 mins ago
Again, you don't need B.E. - the limit is $3^n+1$.
– metamorphy
15 mins ago