Mixing
Evaluate $limlimits_{nto infty} left( cos(1/n)-sin(1/n) right) ^n $?
$begingroup$
How to calculate
$limlimits_{nto infty} left( cos(1/n)-sin(1/n) right) ^n $?
Since $limlimits_{nto infty} frac {cos(1/n)-sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $frac{1}{e}$, but since the form $(to 1)^{to infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
calculus limits
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
$endgroup$
add a comment |
$begingroup$
How to calculate
$limlimits_{nto infty} left( cos(1/n)-sin(1/n) right) ^n $?
Since $limlimits_{nto infty} frac {cos(1/n)-sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $frac{1}{e}$, but since the form $(to 1)^{to infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
calculus limits
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
$endgroup$
$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17
add a comment |
$begingroup$
How to calculate
$limlimits_{nto infty} left( cos(1/n)-sin(1/n) right) ^n $?
Since $limlimits_{nto infty} frac {cos(1/n)-sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $frac{1}{e}$, but since the form $(to 1)^{to infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
calculus limits
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
$endgroup$
How to calculate
$limlimits_{nto infty} left( cos(1/n)-sin(1/n) right) ^n $?
Since $limlimits_{nto infty} frac {cos(1/n)-sin(1/n) }{1-1/n} = 1 $, I guess that the limit above is $frac{1}{e}$, but since the form $(to 1)^{to infty}$ is indeterminate, I don't know how to prove it formally.
Thanks!
calculus limits
calculus limits
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
edited Jan 30 at 19:58
Paramanand Singh
51.2k558170
51.2k558170
asked Jan 30 at 12:34
Dr. JohnDr. John
132
asked Jan 30 at 12:34
Dr. JohnDr. John
132
asked Jan 30 at 12:34
Dr. JohnDr. John
132
132
$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17
add a comment |
$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17
$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17
$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Use $$logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]^n = nlogleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right] = frac{logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]}{frac{1}{n}}$$
and then apply l'Hospital's rule.
answered Jan 30 at 12:44
KlausKlaus
2,955214
$endgroup$
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
add a comment |
$begingroup$
You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$frac1n=dfrac{dfrac1n}{sindfrac1n}sindfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=sindfrac1n$, we have
$$lim_{tto0}left(sqrt{1-t^2}-tright)^{1/t}.$$
Using Taylor,
$$sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$lim_{tto0}left(1-tright)^{1/t}.$$
Or by L'Hospital on the logarithm,
$$frac{log(sqrt{1-t^2}-t)}ttofrac{-dfrac t{sqrt{1-t^2}}-1}{sqrt{1-t^2}-t}to-1.$$
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
Since we recall that
$$sin x=x-{frac {x^{3}}{3!}}+{frac {x^{5}}{5!}}-cdots $$
and
$$cos x=1-{frac {x^{2}}{2!}}+{frac {x^{4}}{4!}}-cdots$$
for all $x$, then
$$cosleft(frac1nright)-sinleft(frac1nright)=1-frac1n+dots$$
and so
$$limlimits_{nto infty} left(cosleft(frac1nright)-sinleft(frac1nright) right) ^n=limlimits_{nto infty} left(1- frac1n right) ^n=e^{-1}, .$$
answered Jan 30 at 13:51
MarkMark
3,46151947
$endgroup$
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
add a comment |
$begingroup$
We have :
$limlimits_{nto infty} left( cos(frac{1}{n})-sin(frac{1}{n}) right) ^n$$=limlimits_{nto infty}(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^n$$=limlimits_{nto infty}left[(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^{frac{1}{cos(frac{1}{n})-sin(frac{1}{n})=1}}right]^{n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$$=e^{limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$
But $limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)$$=limlimits_{nto infty}frac{cos(frac{1}{n})-sin(frac{1}{n})-1}{frac{1}{n}}$$=limlimits_{xto 0}frac{cos x-sin x-1}{x}$$=$$limlimits_{xto 0}(-sin x-cos x)=-1$,so your limit equals $frac{1}{e}$.
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
$begingroup$
You may consider
$left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}$ for $xto 0$ and use
$lim_{tto 0} (1+t)^{frac{1}{t}}= e$
begin{eqnarray*} left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}
& = & left( underbrace{left(1 + frac{cos x - sin x -1 +x}{1-x}right)^{frac{1-x}{cos x - sin x -1 +x}}}_{stackrel{xto 0}{longrightarrow}e}right)^{underbrace{frac{cos x - sin x -1 +x}{x-x^2}}_{stackrel{L'Hop}{sim}frac{-sin x - cos x +1}{1-2x}stackrel{xto 0}{longrightarrow}0}}\
& stackrel{xto 0}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
$endgroup$
add a comment |
$begingroup$
Using the well known limit $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ it is easy to prove that $$lim_{ntoinfty} left(1-frac{1}{n}right)^n=lim_{ntoinfty}dfrac{1}{left(1+dfrac{1}{n-1}right)^{n-1}}cdotfrac{n-1}{n}=frac{1}{e}tag{2}$$ Next we use the following lemma :
Lemma: If ${a_n} $ is sequence such that $n(a_n-1)to 0$ then $a_n^nto 1$.
We can now choose $$a_n=dfrac{cosleft(dfrac{1}{n}right)-sinleft(dfrac{1}{n}right)} {1-dfrac{1}{n}} $$ and note that $$n(a_n-1)=nleft(dfrac{ncosleft(dfrac{1}{n}right)-nsinleft(dfrac{1}{n}right)-n+1} {n-1}right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^nto 1$ and therefore using $(2)$ the desired limit is $1/e$.
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
$endgroup$
add a comment |
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6 Answers
6
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votes
6 Answers
6
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$begingroup$
Use $$logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]^n = nlogleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right] = frac{logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]}{frac{1}{n}}$$
and then apply l'Hospital's rule.
answered Jan 30 at 12:44
KlausKlaus
2,955214
$endgroup$
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
add a comment |
$begingroup$
Use $$logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]^n = nlogleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right] = frac{logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]}{frac{1}{n}}$$
and then apply l'Hospital's rule.
answered Jan 30 at 12:44
KlausKlaus
2,955214
$endgroup$
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
add a comment |
$begingroup$
Use $$logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]^n = nlogleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right] = frac{logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]}{frac{1}{n}}$$
and then apply l'Hospital's rule.
answered Jan 30 at 12:44
KlausKlaus
2,955214
$endgroup$
Use $$logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]^n = nlogleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right] = frac{logleft[cosleft(frac{1}{n}right)-sinleft(frac{1}{n}right)right]}{frac{1}{n}}$$
and then apply l'Hospital's rule.
answered Jan 30 at 12:44
KlausKlaus
2,955214
answered Jan 30 at 12:44
KlausKlaus
2,955214
answered Jan 30 at 12:44
KlausKlaus
2,955214
answered Jan 30 at 12:44
KlausKlaus
2,955214
2,955214
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
add a comment |
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
$begingroup$
You must first prove that the inner limit is not zero - otherwise the function is undefined at infinity.
$endgroup$
– Lucas Henrique
Jan 30 at 13:57
add a comment |
$begingroup$
You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$frac1n=dfrac{dfrac1n}{sindfrac1n}sindfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=sindfrac1n$, we have
$$lim_{tto0}left(sqrt{1-t^2}-tright)^{1/t}.$$
Using Taylor,
$$sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$lim_{tto0}left(1-tright)^{1/t}.$$
Or by L'Hospital on the logarithm,
$$frac{log(sqrt{1-t^2}-t)}ttofrac{-dfrac t{sqrt{1-t^2}}-1}{sqrt{1-t^2}-t}to-1.$$
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$frac1n=dfrac{dfrac1n}{sindfrac1n}sindfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=sindfrac1n$, we have
$$lim_{tto0}left(sqrt{1-t^2}-tright)^{1/t}.$$
Using Taylor,
$$sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$lim_{tto0}left(1-tright)^{1/t}.$$
Or by L'Hospital on the logarithm,
$$frac{log(sqrt{1-t^2}-t)}ttofrac{-dfrac t{sqrt{1-t^2}}-1}{sqrt{1-t^2}-t}to-1.$$
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$frac1n=dfrac{dfrac1n}{sindfrac1n}sindfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=sindfrac1n$, we have
$$lim_{tto0}left(sqrt{1-t^2}-tright)^{1/t}.$$
Using Taylor,
$$sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$lim_{tto0}left(1-tright)^{1/t}.$$
Or by L'Hospital on the logarithm,
$$frac{log(sqrt{1-t^2}-t)}ttofrac{-dfrac t{sqrt{1-t^2}}-1}{sqrt{1-t^2}-t}to-1.$$
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
$endgroup$
You can get rid of the trigonometric functions by rewriting the exponent $n$ using
$$frac1n=dfrac{dfrac1n}{sindfrac1n}sindfrac1n.$$
As the larger fraction tends to $1$, it can be ignored.
Now, with $t:=sindfrac1n$, we have
$$lim_{tto0}left(sqrt{1-t^2}-tright)^{1/t}.$$
Using Taylor,
$$sqrt{1-t^2}-t=1-t+o(t)$$ and the limit is the same as
$$lim_{tto0}left(1-tright)^{1/t}.$$
Or by L'Hospital on the logarithm,
$$frac{log(sqrt{1-t^2}-t)}ttofrac{-dfrac t{sqrt{1-t^2}}-1}{sqrt{1-t^2}-t}to-1.$$
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
edited Jan 30 at 13:39
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
answered Jan 30 at 13:33
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
$begingroup$
Since we recall that
$$sin x=x-{frac {x^{3}}{3!}}+{frac {x^{5}}{5!}}-cdots $$
and
$$cos x=1-{frac {x^{2}}{2!}}+{frac {x^{4}}{4!}}-cdots$$
for all $x$, then
$$cosleft(frac1nright)-sinleft(frac1nright)=1-frac1n+dots$$
and so
$$limlimits_{nto infty} left(cosleft(frac1nright)-sinleft(frac1nright) right) ^n=limlimits_{nto infty} left(1- frac1n right) ^n=e^{-1}, .$$
answered Jan 30 at 13:51
MarkMark
3,46151947
$endgroup$
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
add a comment |
$begingroup$
Since we recall that
$$sin x=x-{frac {x^{3}}{3!}}+{frac {x^{5}}{5!}}-cdots $$
and
$$cos x=1-{frac {x^{2}}{2!}}+{frac {x^{4}}{4!}}-cdots$$
for all $x$, then
$$cosleft(frac1nright)-sinleft(frac1nright)=1-frac1n+dots$$
and so
$$limlimits_{nto infty} left(cosleft(frac1nright)-sinleft(frac1nright) right) ^n=limlimits_{nto infty} left(1- frac1n right) ^n=e^{-1}, .$$
answered Jan 30 at 13:51
MarkMark
3,46151947
$endgroup$
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
add a comment |
$begingroup$
Since we recall that
$$sin x=x-{frac {x^{3}}{3!}}+{frac {x^{5}}{5!}}-cdots $$
and
$$cos x=1-{frac {x^{2}}{2!}}+{frac {x^{4}}{4!}}-cdots$$
for all $x$, then
$$cosleft(frac1nright)-sinleft(frac1nright)=1-frac1n+dots$$
and so
$$limlimits_{nto infty} left(cosleft(frac1nright)-sinleft(frac1nright) right) ^n=limlimits_{nto infty} left(1- frac1n right) ^n=e^{-1}, .$$
answered Jan 30 at 13:51
MarkMark
3,46151947
$endgroup$
Since we recall that
$$sin x=x-{frac {x^{3}}{3!}}+{frac {x^{5}}{5!}}-cdots $$
and
$$cos x=1-{frac {x^{2}}{2!}}+{frac {x^{4}}{4!}}-cdots$$
for all $x$, then
$$cosleft(frac1nright)-sinleft(frac1nright)=1-frac1n+dots$$
and so
$$limlimits_{nto infty} left(cosleft(frac1nright)-sinleft(frac1nright) right) ^n=limlimits_{nto infty} left(1- frac1n right) ^n=e^{-1}, .$$
answered Jan 30 at 13:51
MarkMark
3,46151947
answered Jan 30 at 13:51
MarkMark
3,46151947
answered Jan 30 at 13:51
MarkMark
3,46151947
answered Jan 30 at 13:51
MarkMark
3,46151947
3,46151947
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
add a comment |
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
$begingroup$
Thanks a lot!!!
$endgroup$
– Dr. John
Jan 31 at 20:13
add a comment |
$begingroup$
We have :
$limlimits_{nto infty} left( cos(frac{1}{n})-sin(frac{1}{n}) right) ^n$$=limlimits_{nto infty}(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^n$$=limlimits_{nto infty}left[(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^{frac{1}{cos(frac{1}{n})-sin(frac{1}{n})=1}}right]^{n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$$=e^{limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$
But $limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)$$=limlimits_{nto infty}frac{cos(frac{1}{n})-sin(frac{1}{n})-1}{frac{1}{n}}$$=limlimits_{xto 0}frac{cos x-sin x-1}{x}$$=$$limlimits_{xto 0}(-sin x-cos x)=-1$,so your limit equals $frac{1}{e}$.
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
$begingroup$
We have :
$limlimits_{nto infty} left( cos(frac{1}{n})-sin(frac{1}{n}) right) ^n$$=limlimits_{nto infty}(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^n$$=limlimits_{nto infty}left[(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^{frac{1}{cos(frac{1}{n})-sin(frac{1}{n})=1}}right]^{n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$$=e^{limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$
But $limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)$$=limlimits_{nto infty}frac{cos(frac{1}{n})-sin(frac{1}{n})-1}{frac{1}{n}}$$=limlimits_{xto 0}frac{cos x-sin x-1}{x}$$=$$limlimits_{xto 0}(-sin x-cos x)=-1$,so your limit equals $frac{1}{e}$.
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
$endgroup$
add a comment |
$begingroup$
We have :
$limlimits_{nto infty} left( cos(frac{1}{n})-sin(frac{1}{n}) right) ^n$$=limlimits_{nto infty}(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^n$$=limlimits_{nto infty}left[(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^{frac{1}{cos(frac{1}{n})-sin(frac{1}{n})=1}}right]^{n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$$=e^{limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$
But $limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)$$=limlimits_{nto infty}frac{cos(frac{1}{n})-sin(frac{1}{n})-1}{frac{1}{n}}$$=limlimits_{xto 0}frac{cos x-sin x-1}{x}$$=$$limlimits_{xto 0}(-sin x-cos x)=-1$,so your limit equals $frac{1}{e}$.
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
$endgroup$
We have :
$limlimits_{nto infty} left( cos(frac{1}{n})-sin(frac{1}{n}) right) ^n$$=limlimits_{nto infty}(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^n$$=limlimits_{nto infty}left[(1+cos(frac{1}{n})-sin(frac{1}{n})-1)^{frac{1}{cos(frac{1}{n})-sin(frac{1}{n})=1}}right]^{n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$$=e^{limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)}$
But $limlimits_{nto infty}n(cos(frac{1}{n})-sin(frac{1}{n})-1)$$=limlimits_{nto infty}frac{cos(frac{1}{n})-sin(frac{1}{n})-1}{frac{1}{n}}$$=limlimits_{xto 0}frac{cos x-sin x-1}{x}$$=$$limlimits_{xto 0}(-sin x-cos x)=-1$,so your limit equals $frac{1}{e}$.
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
answered Jan 30 at 21:57
AlexdanutAlexdanut
1738
1738
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$begingroup$
You may consider
$left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}$ for $xto 0$ and use
$lim_{tto 0} (1+t)^{frac{1}{t}}= e$
begin{eqnarray*} left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}
& = & left( underbrace{left(1 + frac{cos x - sin x -1 +x}{1-x}right)^{frac{1-x}{cos x - sin x -1 +x}}}_{stackrel{xto 0}{longrightarrow}e}right)^{underbrace{frac{cos x - sin x -1 +x}{x-x^2}}_{stackrel{L'Hop}{sim}frac{-sin x - cos x +1}{1-2x}stackrel{xto 0}{longrightarrow}0}}\
& stackrel{xto 0}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
$endgroup$
add a comment |
$begingroup$
You may consider
$left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}$ for $xto 0$ and use
$lim_{tto 0} (1+t)^{frac{1}{t}}= e$
begin{eqnarray*} left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}
& = & left( underbrace{left(1 + frac{cos x - sin x -1 +x}{1-x}right)^{frac{1-x}{cos x - sin x -1 +x}}}_{stackrel{xto 0}{longrightarrow}e}right)^{underbrace{frac{cos x - sin x -1 +x}{x-x^2}}_{stackrel{L'Hop}{sim}frac{-sin x - cos x +1}{1-2x}stackrel{xto 0}{longrightarrow}0}}\
& stackrel{xto 0}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
$endgroup$
add a comment |
$begingroup$
You may consider
$left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}$ for $xto 0$ and use
$lim_{tto 0} (1+t)^{frac{1}{t}}= e$
begin{eqnarray*} left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}
& = & left( underbrace{left(1 + frac{cos x - sin x -1 +x}{1-x}right)^{frac{1-x}{cos x - sin x -1 +x}}}_{stackrel{xto 0}{longrightarrow}e}right)^{underbrace{frac{cos x - sin x -1 +x}{x-x^2}}_{stackrel{L'Hop}{sim}frac{-sin x - cos x +1}{1-2x}stackrel{xto 0}{longrightarrow}0}}\
& stackrel{xto 0}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
$endgroup$
You may consider
$left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}$ for $xto 0$ and use
$lim_{tto 0} (1+t)^{frac{1}{t}}= e$
begin{eqnarray*} left(frac{cos x - sin x}{1-x}right)^{frac{1}{x}}
& = & left( underbrace{left(1 + frac{cos x - sin x -1 +x}{1-x}right)^{frac{1-x}{cos x - sin x -1 +x}}}_{stackrel{xto 0}{longrightarrow}e}right)^{underbrace{frac{cos x - sin x -1 +x}{x-x^2}}_{stackrel{L'Hop}{sim}frac{-sin x - cos x +1}{1-2x}stackrel{xto 0}{longrightarrow}0}}\
& stackrel{xto 0}{longrightarrow} & e^0 = 1
end{eqnarray*}
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
answered Jan 30 at 13:23
trancelocationtrancelocation
13.6k1828
13.6k1828
add a comment |
add a comment |
$begingroup$
Using the well known limit $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ it is easy to prove that $$lim_{ntoinfty} left(1-frac{1}{n}right)^n=lim_{ntoinfty}dfrac{1}{left(1+dfrac{1}{n-1}right)^{n-1}}cdotfrac{n-1}{n}=frac{1}{e}tag{2}$$ Next we use the following lemma :
Lemma: If ${a_n} $ is sequence such that $n(a_n-1)to 0$ then $a_n^nto 1$.
We can now choose $$a_n=dfrac{cosleft(dfrac{1}{n}right)-sinleft(dfrac{1}{n}right)} {1-dfrac{1}{n}} $$ and note that $$n(a_n-1)=nleft(dfrac{ncosleft(dfrac{1}{n}right)-nsinleft(dfrac{1}{n}right)-n+1} {n-1}right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^nto 1$ and therefore using $(2)$ the desired limit is $1/e$.
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
$endgroup$
add a comment |
$begingroup$
Using the well known limit $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ it is easy to prove that $$lim_{ntoinfty} left(1-frac{1}{n}right)^n=lim_{ntoinfty}dfrac{1}{left(1+dfrac{1}{n-1}right)^{n-1}}cdotfrac{n-1}{n}=frac{1}{e}tag{2}$$ Next we use the following lemma :
Lemma: If ${a_n} $ is sequence such that $n(a_n-1)to 0$ then $a_n^nto 1$.
We can now choose $$a_n=dfrac{cosleft(dfrac{1}{n}right)-sinleft(dfrac{1}{n}right)} {1-dfrac{1}{n}} $$ and note that $$n(a_n-1)=nleft(dfrac{ncosleft(dfrac{1}{n}right)-nsinleft(dfrac{1}{n}right)-n+1} {n-1}right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^nto 1$ and therefore using $(2)$ the desired limit is $1/e$.
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
$endgroup$
add a comment |
$begingroup$
Using the well known limit $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ it is easy to prove that $$lim_{ntoinfty} left(1-frac{1}{n}right)^n=lim_{ntoinfty}dfrac{1}{left(1+dfrac{1}{n-1}right)^{n-1}}cdotfrac{n-1}{n}=frac{1}{e}tag{2}$$ Next we use the following lemma :
Lemma: If ${a_n} $ is sequence such that $n(a_n-1)to 0$ then $a_n^nto 1$.
We can now choose $$a_n=dfrac{cosleft(dfrac{1}{n}right)-sinleft(dfrac{1}{n}right)} {1-dfrac{1}{n}} $$ and note that $$n(a_n-1)=nleft(dfrac{ncosleft(dfrac{1}{n}right)-nsinleft(dfrac{1}{n}right)-n+1} {n-1}right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^nto 1$ and therefore using $(2)$ the desired limit is $1/e$.
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
$endgroup$
Using the well known limit $$lim_{ntoinfty} left(1+frac{1}{n}right)^n=etag{1}$$ it is easy to prove that $$lim_{ntoinfty} left(1-frac{1}{n}right)^n=lim_{ntoinfty}dfrac{1}{left(1+dfrac{1}{n-1}right)^{n-1}}cdotfrac{n-1}{n}=frac{1}{e}tag{2}$$ Next we use the following lemma :
Lemma: If ${a_n} $ is sequence such that $n(a_n-1)to 0$ then $a_n^nto 1$.
We can now choose $$a_n=dfrac{cosleft(dfrac{1}{n}right)-sinleft(dfrac{1}{n}right)} {1-dfrac{1}{n}} $$ and note that $$n(a_n-1)=nleft(dfrac{ncosleft(dfrac{1}{n}right)-nsinleft(dfrac{1}{n}right)-n+1} {n-1}right) $$ and the above clearly tends to $0$ so that by lemma above $a_n^nto 1$ and therefore using $(2)$ the desired limit is $1/e$.
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
answered Jan 30 at 15:34
Paramanand SinghParamanand Singh
51.2k558170
51.2k558170
add a comment |
add a comment |
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$begingroup$
You are right as $cos1/n-sin1/n=1-1/n+o(1/n)$.
$endgroup$
– Yves Daoust
Jan 30 at 13:17