Mixing
Idea for $limlimits_{x to frac{pi}{2}} left( tan left( tfrac{pi}{4} sin xright)right)^{1/ ( tan (pi sin x))}$
$begingroup$
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
limits trigonometry limits-without-lhopital
edited Jan 26 at 17:43
rtybase
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
$endgroup$
|
show 3 more comments
$begingroup$
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
limits trigonometry limits-without-lhopital
edited Jan 26 at 17:43
rtybase
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
$endgroup$
1
$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
1
$begingroup$
tangent then exp
$endgroup$
– bono95zg
Jan 25 at 19:05
|
show 3 more comments
$begingroup$
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
limits trigonometry limits-without-lhopital
edited Jan 26 at 17:43
rtybase
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
$endgroup$
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right).$$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.
ps. can't use lhopital
limits trigonometry limits-without-lhopital
limits trigonometry limits-without-lhopital
edited Jan 26 at 17:43
rtybase
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
edited Jan 26 at 17:43
rtybase
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
edited Jan 26 at 17:43
rtybase
11.5k31534
edited Jan 26 at 17:43
rtybase
11.5k31534
edited Jan 26 at 17:43
rtybase
11.5k31534
11.5k31534
asked Jan 25 at 18:40
bono95zgbono95zg
112
asked Jan 25 at 18:40
bono95zgbono95zg
112
asked Jan 25 at 18:40
bono95zgbono95zg
112
112
1
$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
1
$begingroup$
tangent then exp
$endgroup$
– bono95zg
Jan 25 at 19:05
|
show 3 more comments
1
$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
1
$begingroup$
tangent then exp
$endgroup$
– bono95zg
Jan 25 at 19:05
1
1
$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
1
1
$begingroup$
tangent then exp
$endgroup$
– bono95zg
Jan 25 at 19:05
$begingroup$
tangent then exp
$endgroup$
– bono95zg
Jan 25 at 19:05
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Warning: you'll want to double-check all my arithmetic.
With $y:=pisin x$ we can rewrite this as $explim_{ytopi}frac{lntanfrac{y}{4}}{tan y}$. Define $t:=tanfrac{y}{4}$ so $$tanfrac{y}{2}=frac{2t}{1-t^2},,tan y=frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$explim_{tto 1}frac{(1-6t^2+t^4)ln t}{4t(1-t^2)}=exp-lim_{tto 1}frac{ln t}{t(1-t^2)}.$$Finally, write $t=1+epsilon$ so the limit is$$explim_{epsilonto 0}frac{ln(1+epsilon)}{(1+epsilon)epsilon(2+epsilon)}=explim_{epsilonto 0}frac{ln(1+epsilon)}{2epsilon}=sqrt{e}.$$
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
add a comment |
$begingroup$
You might first do the substitution $t=frac{pi}{4}sin x$ that brings the limit into the form
$$
lim_{ttopi/4^-}(tan t)^{1/tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
lim_{ttopi/4^-}frac{logtan t}{tan 4t}
$$
Apply $t=pi/4-u$,
$$
tan t=frac{1-tan u}{1+tan u}
$$
and $tan4t=tan(pi-4u)=-tan 4u$. Therefore the limit becomes
$$
lim_{uto0^+}frac{log(1+tan u)-log(1-tan u)}{tan 4u}
$$
Now, for instance,
$$
lim_{uto0^+}frac{log(1+tan u)}{tan4u}=
lim_{uto0^+}frac{log(1+tan u)}{tan u}frac{tan u}{u}frac{4u}{tan 4u}frac{1}{4}
$$
answered Jan 25 at 21:54
egregegreg
184k1486206
$endgroup$
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
add a comment |
$begingroup$
Alternatively:
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right)=\
lim_{x to frac{pi}{2}} left[1+tan left(frac{pi}{4}sin xright)-1right]^left({dfrac 1{tan(pi sin x)}}right)=\
exp left[lim_{x to frac{pi}{2}} frac{tan left(frac{pi}{4}sin xright)-1}{tan(pi sin x)} right]=\
exp left[lim_{x to frac{pi}{2}} frac{sin left(frac{pi}{4}sin xright)-cos left(frac{pi}{4}sin xright)}{sin(pi sin x)}cdot frac{cos (pi sin x)}{cos left(frac{pi}{4}sin xright)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{cos left(frac{pi}{2}sin xright)}{sin(pi sin x)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{1}{2sin(frac{pi}{2} sin x)}right] =exp left[frac12right]=sqrt{e}.
$$
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
$endgroup$
add a comment |
$begingroup$
Let $dfrac{pisin x}4=y$
$$lim_{ytofracpi4}(tan y)^{frac1{tan4y}}=left[lim_{ytofracpi4}(1+tan y-1)^{frac1{tan y -1}}right]^{lim_{ytofracpi4}dfrac{tan y-1}{tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= lim_{ytofracpi4}dfrac{tan y-1}{tan4y}=lim_{ytofracpi4}dfrac{tan y-tandfracpi4}{tan4y-tanpi}$
Method$#1:$
$$F=dfrac{dfrac{d(tan y)}{dy}_{(text { at }y=pi/4)}}{dfrac{d(tan4y)}{dy}_{(text { at }y=pi/4)}}=?$$
Method$#2:$
$$F=lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)cos4y}{sin4ycos ycosdfracpi4}$$
$$=sqrt2lim_{ytofracpi4}dfrac{cos4y}{cos y}cdotdfrac{lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)}{left(y-dfracpi4right)}}{-4lim_{ytofracpi4}dfrac{sin4left(y-dfracpi4right)}{4left(y-dfracpi4right)}}text{ as }sin(4y-pi)=-sin(pi-4y)=-sin4y$$
$$=dfrac{sqrt2cospi}{-4cosdfracpi4}=?$$
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
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oldest
votes
$begingroup$
Warning: you'll want to double-check all my arithmetic.
With $y:=pisin x$ we can rewrite this as $explim_{ytopi}frac{lntanfrac{y}{4}}{tan y}$. Define $t:=tanfrac{y}{4}$ so $$tanfrac{y}{2}=frac{2t}{1-t^2},,tan y=frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$explim_{tto 1}frac{(1-6t^2+t^4)ln t}{4t(1-t^2)}=exp-lim_{tto 1}frac{ln t}{t(1-t^2)}.$$Finally, write $t=1+epsilon$ so the limit is$$explim_{epsilonto 0}frac{ln(1+epsilon)}{(1+epsilon)epsilon(2+epsilon)}=explim_{epsilonto 0}frac{ln(1+epsilon)}{2epsilon}=sqrt{e}.$$
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
add a comment |
$begingroup$
Warning: you'll want to double-check all my arithmetic.
With $y:=pisin x$ we can rewrite this as $explim_{ytopi}frac{lntanfrac{y}{4}}{tan y}$. Define $t:=tanfrac{y}{4}$ so $$tanfrac{y}{2}=frac{2t}{1-t^2},,tan y=frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$explim_{tto 1}frac{(1-6t^2+t^4)ln t}{4t(1-t^2)}=exp-lim_{tto 1}frac{ln t}{t(1-t^2)}.$$Finally, write $t=1+epsilon$ so the limit is$$explim_{epsilonto 0}frac{ln(1+epsilon)}{(1+epsilon)epsilon(2+epsilon)}=explim_{epsilonto 0}frac{ln(1+epsilon)}{2epsilon}=sqrt{e}.$$
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
$endgroup$
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
add a comment |
$begingroup$
Warning: you'll want to double-check all my arithmetic.
With $y:=pisin x$ we can rewrite this as $explim_{ytopi}frac{lntanfrac{y}{4}}{tan y}$. Define $t:=tanfrac{y}{4}$ so $$tanfrac{y}{2}=frac{2t}{1-t^2},,tan y=frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$explim_{tto 1}frac{(1-6t^2+t^4)ln t}{4t(1-t^2)}=exp-lim_{tto 1}frac{ln t}{t(1-t^2)}.$$Finally, write $t=1+epsilon$ so the limit is$$explim_{epsilonto 0}frac{ln(1+epsilon)}{(1+epsilon)epsilon(2+epsilon)}=explim_{epsilonto 0}frac{ln(1+epsilon)}{2epsilon}=sqrt{e}.$$
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
$endgroup$
Warning: you'll want to double-check all my arithmetic.
With $y:=pisin x$ we can rewrite this as $explim_{ytopi}frac{lntanfrac{y}{4}}{tan y}$. Define $t:=tanfrac{y}{4}$ so $$tanfrac{y}{2}=frac{2t}{1-t^2},,tan y=frac{4t(1-t^2)}{1-6t^2+t^4}.$$Then your limit is $$explim_{tto 1}frac{(1-6t^2+t^4)ln t}{4t(1-t^2)}=exp-lim_{tto 1}frac{ln t}{t(1-t^2)}.$$Finally, write $t=1+epsilon$ so the limit is$$explim_{epsilonto 0}frac{ln(1+epsilon)}{(1+epsilon)epsilon(2+epsilon)}=explim_{epsilonto 0}frac{ln(1+epsilon)}{2epsilon}=sqrt{e}.$$
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
answered Jan 25 at 19:07
J.G.J.G.
30.9k23149
30.9k23149
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
add a comment |
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
$begingroup$
looks correct. thank you.
$endgroup$
– bono95zg
Jan 25 at 19:31
add a comment |
$begingroup$
You might first do the substitution $t=frac{pi}{4}sin x$ that brings the limit into the form
$$
lim_{ttopi/4^-}(tan t)^{1/tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
lim_{ttopi/4^-}frac{logtan t}{tan 4t}
$$
Apply $t=pi/4-u$,
$$
tan t=frac{1-tan u}{1+tan u}
$$
and $tan4t=tan(pi-4u)=-tan 4u$. Therefore the limit becomes
$$
lim_{uto0^+}frac{log(1+tan u)-log(1-tan u)}{tan 4u}
$$
Now, for instance,
$$
lim_{uto0^+}frac{log(1+tan u)}{tan4u}=
lim_{uto0^+}frac{log(1+tan u)}{tan u}frac{tan u}{u}frac{4u}{tan 4u}frac{1}{4}
$$
answered Jan 25 at 21:54
egregegreg
184k1486206
$endgroup$
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
add a comment |
$begingroup$
You might first do the substitution $t=frac{pi}{4}sin x$ that brings the limit into the form
$$
lim_{ttopi/4^-}(tan t)^{1/tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
lim_{ttopi/4^-}frac{logtan t}{tan 4t}
$$
Apply $t=pi/4-u$,
$$
tan t=frac{1-tan u}{1+tan u}
$$
and $tan4t=tan(pi-4u)=-tan 4u$. Therefore the limit becomes
$$
lim_{uto0^+}frac{log(1+tan u)-log(1-tan u)}{tan 4u}
$$
Now, for instance,
$$
lim_{uto0^+}frac{log(1+tan u)}{tan4u}=
lim_{uto0^+}frac{log(1+tan u)}{tan u}frac{tan u}{u}frac{4u}{tan 4u}frac{1}{4}
$$
answered Jan 25 at 21:54
egregegreg
184k1486206
$endgroup$
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
add a comment |
$begingroup$
You might first do the substitution $t=frac{pi}{4}sin x$ that brings the limit into the form
$$
lim_{ttopi/4^-}(tan t)^{1/tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
lim_{ttopi/4^-}frac{logtan t}{tan 4t}
$$
Apply $t=pi/4-u$,
$$
tan t=frac{1-tan u}{1+tan u}
$$
and $tan4t=tan(pi-4u)=-tan 4u$. Therefore the limit becomes
$$
lim_{uto0^+}frac{log(1+tan u)-log(1-tan u)}{tan 4u}
$$
Now, for instance,
$$
lim_{uto0^+}frac{log(1+tan u)}{tan4u}=
lim_{uto0^+}frac{log(1+tan u)}{tan u}frac{tan u}{u}frac{4u}{tan 4u}frac{1}{4}
$$
answered Jan 25 at 21:54
egregegreg
184k1486206
$endgroup$
You might first do the substitution $t=frac{pi}{4}sin x$ that brings the limit into the form
$$
lim_{ttopi/4^-}(tan t)^{1/tan4t}
$$
(prove first that both the limit from the left and from the right lead to the same limit).
Now it's better to pass to the logarithm:
$$
lim_{ttopi/4^-}frac{logtan t}{tan 4t}
$$
Apply $t=pi/4-u$,
$$
tan t=frac{1-tan u}{1+tan u}
$$
and $tan4t=tan(pi-4u)=-tan 4u$. Therefore the limit becomes
$$
lim_{uto0^+}frac{log(1+tan u)-log(1-tan u)}{tan 4u}
$$
Now, for instance,
$$
lim_{uto0^+}frac{log(1+tan u)}{tan4u}=
lim_{uto0^+}frac{log(1+tan u)}{tan u}frac{tan u}{u}frac{4u}{tan 4u}frac{1}{4}
$$
answered Jan 25 at 21:54
egregegreg
184k1486206
answered Jan 25 at 21:54
egregegreg
184k1486206
answered Jan 25 at 21:54
egregegreg
184k1486206
answered Jan 25 at 21:54
egregegreg
184k1486206
184k1486206
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
add a comment |
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
If you consider $1pmtan u$, you can do both calculations for the price of one.
$endgroup$
– J.G.
Jan 25 at 22:20
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
$begingroup$
@J.G. That's indeed the idea: the second limit to compute is essentially the same.
$endgroup$
– egreg
Jan 25 at 22:27
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I know; I was making an edit suggestion.
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– J.G.
Jan 25 at 22:45
$begingroup$
I know; I was making an edit suggestion.
$endgroup$
– J.G.
Jan 25 at 22:45
add a comment |
$begingroup$
Alternatively:
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right)=\
lim_{x to frac{pi}{2}} left[1+tan left(frac{pi}{4}sin xright)-1right]^left({dfrac 1{tan(pi sin x)}}right)=\
exp left[lim_{x to frac{pi}{2}} frac{tan left(frac{pi}{4}sin xright)-1}{tan(pi sin x)} right]=\
exp left[lim_{x to frac{pi}{2}} frac{sin left(frac{pi}{4}sin xright)-cos left(frac{pi}{4}sin xright)}{sin(pi sin x)}cdot frac{cos (pi sin x)}{cos left(frac{pi}{4}sin xright)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{cos left(frac{pi}{2}sin xright)}{sin(pi sin x)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{1}{2sin(frac{pi}{2} sin x)}right] =exp left[frac12right]=sqrt{e}.
$$
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
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add a comment |
$begingroup$
Alternatively:
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right)=\
lim_{x to frac{pi}{2}} left[1+tan left(frac{pi}{4}sin xright)-1right]^left({dfrac 1{tan(pi sin x)}}right)=\
exp left[lim_{x to frac{pi}{2}} frac{tan left(frac{pi}{4}sin xright)-1}{tan(pi sin x)} right]=\
exp left[lim_{x to frac{pi}{2}} frac{sin left(frac{pi}{4}sin xright)-cos left(frac{pi}{4}sin xright)}{sin(pi sin x)}cdot frac{cos (pi sin x)}{cos left(frac{pi}{4}sin xright)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{cos left(frac{pi}{2}sin xright)}{sin(pi sin x)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{1}{2sin(frac{pi}{2} sin x)}right] =exp left[frac12right]=sqrt{e}.
$$
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
$endgroup$
add a comment |
$begingroup$
Alternatively:
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right)=\
lim_{x to frac{pi}{2}} left[1+tan left(frac{pi}{4}sin xright)-1right]^left({dfrac 1{tan(pi sin x)}}right)=\
exp left[lim_{x to frac{pi}{2}} frac{tan left(frac{pi}{4}sin xright)-1}{tan(pi sin x)} right]=\
exp left[lim_{x to frac{pi}{2}} frac{sin left(frac{pi}{4}sin xright)-cos left(frac{pi}{4}sin xright)}{sin(pi sin x)}cdot frac{cos (pi sin x)}{cos left(frac{pi}{4}sin xright)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{cos left(frac{pi}{2}sin xright)}{sin(pi sin x)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{1}{2sin(frac{pi}{2} sin x)}right] =exp left[frac12right]=sqrt{e}.
$$
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
$endgroup$
Alternatively:
$$lim_{x to frac{pi}{2}} tan left(frac{pi}{4}sin xright)^left({dfrac 1{tan(pi sin x)}}right)=\
lim_{x to frac{pi}{2}} left[1+tan left(frac{pi}{4}sin xright)-1right]^left({dfrac 1{tan(pi sin x)}}right)=\
exp left[lim_{x to frac{pi}{2}} frac{tan left(frac{pi}{4}sin xright)-1}{tan(pi sin x)} right]=\
exp left[lim_{x to frac{pi}{2}} frac{sin left(frac{pi}{4}sin xright)-cos left(frac{pi}{4}sin xright)}{sin(pi sin x)}cdot frac{cos (pi sin x)}{cos left(frac{pi}{4}sin xright)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{cos left(frac{pi}{2}sin xright)}{sin(pi sin x)}right] =\
exp left[lim_{x to frac{pi}{2}} frac{1}{2sin(frac{pi}{2} sin x)}right] =exp left[frac12right]=sqrt{e}.
$$
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
answered Jan 26 at 13:10
farruhotafarruhota
21.3k2841
21.3k2841
add a comment |
add a comment |
$begingroup$
Let $dfrac{pisin x}4=y$
$$lim_{ytofracpi4}(tan y)^{frac1{tan4y}}=left[lim_{ytofracpi4}(1+tan y-1)^{frac1{tan y -1}}right]^{lim_{ytofracpi4}dfrac{tan y-1}{tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= lim_{ytofracpi4}dfrac{tan y-1}{tan4y}=lim_{ytofracpi4}dfrac{tan y-tandfracpi4}{tan4y-tanpi}$
Method$#1:$
$$F=dfrac{dfrac{d(tan y)}{dy}_{(text { at }y=pi/4)}}{dfrac{d(tan4y)}{dy}_{(text { at }y=pi/4)}}=?$$
Method$#2:$
$$F=lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)cos4y}{sin4ycos ycosdfracpi4}$$
$$=sqrt2lim_{ytofracpi4}dfrac{cos4y}{cos y}cdotdfrac{lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)}{left(y-dfracpi4right)}}{-4lim_{ytofracpi4}dfrac{sin4left(y-dfracpi4right)}{4left(y-dfracpi4right)}}text{ as }sin(4y-pi)=-sin(pi-4y)=-sin4y$$
$$=dfrac{sqrt2cospi}{-4cosdfracpi4}=?$$
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
add a comment |
$begingroup$
Let $dfrac{pisin x}4=y$
$$lim_{ytofracpi4}(tan y)^{frac1{tan4y}}=left[lim_{ytofracpi4}(1+tan y-1)^{frac1{tan y -1}}right]^{lim_{ytofracpi4}dfrac{tan y-1}{tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= lim_{ytofracpi4}dfrac{tan y-1}{tan4y}=lim_{ytofracpi4}dfrac{tan y-tandfracpi4}{tan4y-tanpi}$
Method$#1:$
$$F=dfrac{dfrac{d(tan y)}{dy}_{(text { at }y=pi/4)}}{dfrac{d(tan4y)}{dy}_{(text { at }y=pi/4)}}=?$$
Method$#2:$
$$F=lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)cos4y}{sin4ycos ycosdfracpi4}$$
$$=sqrt2lim_{ytofracpi4}dfrac{cos4y}{cos y}cdotdfrac{lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)}{left(y-dfracpi4right)}}{-4lim_{ytofracpi4}dfrac{sin4left(y-dfracpi4right)}{4left(y-dfracpi4right)}}text{ as }sin(4y-pi)=-sin(pi-4y)=-sin4y$$
$$=dfrac{sqrt2cospi}{-4cosdfracpi4}=?$$
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
add a comment |
$begingroup$
Let $dfrac{pisin x}4=y$
$$lim_{ytofracpi4}(tan y)^{frac1{tan4y}}=left[lim_{ytofracpi4}(1+tan y-1)^{frac1{tan y -1}}right]^{lim_{ytofracpi4}dfrac{tan y-1}{tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= lim_{ytofracpi4}dfrac{tan y-1}{tan4y}=lim_{ytofracpi4}dfrac{tan y-tandfracpi4}{tan4y-tanpi}$
Method$#1:$
$$F=dfrac{dfrac{d(tan y)}{dy}_{(text { at }y=pi/4)}}{dfrac{d(tan4y)}{dy}_{(text { at }y=pi/4)}}=?$$
Method$#2:$
$$F=lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)cos4y}{sin4ycos ycosdfracpi4}$$
$$=sqrt2lim_{ytofracpi4}dfrac{cos4y}{cos y}cdotdfrac{lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)}{left(y-dfracpi4right)}}{-4lim_{ytofracpi4}dfrac{sin4left(y-dfracpi4right)}{4left(y-dfracpi4right)}}text{ as }sin(4y-pi)=-sin(pi-4y)=-sin4y$$
$$=dfrac{sqrt2cospi}{-4cosdfracpi4}=?$$
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
$endgroup$
Let $dfrac{pisin x}4=y$
$$lim_{ytofracpi4}(tan y)^{frac1{tan4y}}=left[lim_{ytofracpi4}(1+tan y-1)^{frac1{tan y -1}}right]^{lim_{ytofracpi4}dfrac{tan y-1}{tan4y}}$$
The inner limit converges to $e$
The exponent$(F)= lim_{ytofracpi4}dfrac{tan y-1}{tan4y}=lim_{ytofracpi4}dfrac{tan y-tandfracpi4}{tan4y-tanpi}$
Method$#1:$
$$F=dfrac{dfrac{d(tan y)}{dy}_{(text { at }y=pi/4)}}{dfrac{d(tan4y)}{dy}_{(text { at }y=pi/4)}}=?$$
Method$#2:$
$$F=lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)cos4y}{sin4ycos ycosdfracpi4}$$
$$=sqrt2lim_{ytofracpi4}dfrac{cos4y}{cos y}cdotdfrac{lim_{ytofracpi4}dfrac{sinleft(y-dfracpi4right)}{left(y-dfracpi4right)}}{-4lim_{ytofracpi4}dfrac{sin4left(y-dfracpi4right)}{4left(y-dfracpi4right)}}text{ as }sin(4y-pi)=-sin(pi-4y)=-sin4y$$
$$=dfrac{sqrt2cospi}{-4cosdfracpi4}=?$$
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
edited Jan 26 at 15:11
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
answered Jan 26 at 14:56
lab bhattacharjeelab bhattacharjee
227k15158276
227k15158276
add a comment |
add a comment |
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$begingroup$
Sorry we've had some mishaps converting your title to MathJax. Could you please clarify whether $pisin x/4$ means $(pisin x)/4$ or $pisin (x/4)$?
$endgroup$
– J.G.
Jan 25 at 18:45
$begingroup$
yeah its pi/4 *sinx
$endgroup$
– bono95zg
Jan 25 at 18:46
$begingroup$
(πsinx)/4 to be exact
$endgroup$
– bono95zg
Jan 25 at 18:48
$begingroup$
L'Hopital's rule is not required even, just put in $x=1$. You would need a calculator for evaluation of this limit. Anyhow, the answer is: $$displaystyle tan (dfrac{pi}{4} sin (1))^{cot (pi sin (1))} approx 1.5885$$
$endgroup$
– Paras Khosla
Jan 25 at 18:52
1
$begingroup$
tangent then exp
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– bono95zg
Jan 25 at 19:05