Solving $lim_{xto 0}bigg( Big( frac{1+sin(x)cos(alpha x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$: my...
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$$lim_{xto 0}bigg( Big( frac{1+sin(x)cos(alpha x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( frac{1+sin(x)cos(alpha x)+sin(x)cos(beta x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( 1+frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}Big)^{cot^3(x) cdot frac{1+sin(x)cos(beta x)}{sin(x)cos(alpha x)-sin(x)cos(beta x)} cdot frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}} bigg) $$
$$e^{lim_{xto 0} big( frac{cos^3(x)sin(x)(cos(alpha x) - cos(beta x))}{sin(x)sin^2(x) (1+sin(x)cos(beta x))} big)}$$
Currently disregarding the exponent, because the font is small.
$$lim_{xto 0} big( frac{cos(alpha x) - cos(beta x)}{sin^2(x)} big)$$
Through Taylor's expansion,
$$lim_{xto 0} big( frac{1-frac{alpha^2x^2}{2} -1 + frac{beta^2x^2}{2}}{x^2} big)$$
And the answer is
$$e^{frac{beta^2 - alpha^2}{2}}.$$
The answer in the book, however, is $e^{beta^2 - alpha^2}$. The formula for the difference of two cosines also leaves me with $e^{frac{beta^2 - alpha^2}{2}}$, and I feel like there's a typo, but maybe I went wrong somewhere?
calculus limits proof-verification
edited Nov 18 at 9:22
Kemono Chen
1,709330
asked Nov 18 at 9:08
fragileradius
278114
add a comment |
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0
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$$lim_{xto 0}bigg( Big( frac{1+sin(x)cos(alpha x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( frac{1+sin(x)cos(alpha x)+sin(x)cos(beta x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( 1+frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}Big)^{cot^3(x) cdot frac{1+sin(x)cos(beta x)}{sin(x)cos(alpha x)-sin(x)cos(beta x)} cdot frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}} bigg) $$
$$e^{lim_{xto 0} big( frac{cos^3(x)sin(x)(cos(alpha x) - cos(beta x))}{sin(x)sin^2(x) (1+sin(x)cos(beta x))} big)}$$
Currently disregarding the exponent, because the font is small.
$$lim_{xto 0} big( frac{cos(alpha x) - cos(beta x)}{sin^2(x)} big)$$
Through Taylor's expansion,
$$lim_{xto 0} big( frac{1-frac{alpha^2x^2}{2} -1 + frac{beta^2x^2}{2}}{x^2} big)$$
And the answer is
$$e^{frac{beta^2 - alpha^2}{2}}.$$
The answer in the book, however, is $e^{beta^2 - alpha^2}$. The formula for the difference of two cosines also leaves me with $e^{frac{beta^2 - alpha^2}{2}}$, and I feel like there's a typo, but maybe I went wrong somewhere?
calculus limits proof-verification
edited Nov 18 at 9:22
Kemono Chen
1,709330
asked Nov 18 at 9:08
fragileradius
278114
2
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26
add a comment |
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up vote
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$$lim_{xto 0}bigg( Big( frac{1+sin(x)cos(alpha x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( frac{1+sin(x)cos(alpha x)+sin(x)cos(beta x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( 1+frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}Big)^{cot^3(x) cdot frac{1+sin(x)cos(beta x)}{sin(x)cos(alpha x)-sin(x)cos(beta x)} cdot frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}} bigg) $$
$$e^{lim_{xto 0} big( frac{cos^3(x)sin(x)(cos(alpha x) - cos(beta x))}{sin(x)sin^2(x) (1+sin(x)cos(beta x))} big)}$$
Currently disregarding the exponent, because the font is small.
$$lim_{xto 0} big( frac{cos(alpha x) - cos(beta x)}{sin^2(x)} big)$$
Through Taylor's expansion,
$$lim_{xto 0} big( frac{1-frac{alpha^2x^2}{2} -1 + frac{beta^2x^2}{2}}{x^2} big)$$
And the answer is
$$e^{frac{beta^2 - alpha^2}{2}}.$$
The answer in the book, however, is $e^{beta^2 - alpha^2}$. The formula for the difference of two cosines also leaves me with $e^{frac{beta^2 - alpha^2}{2}}$, and I feel like there's a typo, but maybe I went wrong somewhere?
calculus limits proof-verification
edited Nov 18 at 9:22
Kemono Chen
1,709330
asked Nov 18 at 9:08
fragileradius
278114
$$lim_{xto 0}bigg( Big( frac{1+sin(x)cos(alpha x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( frac{1+sin(x)cos(alpha x)+sin(x)cos(beta x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)} Big) ^{cot^3(x)}bigg)$$
$$lim_{xto 0} bigg( Big( 1+frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}Big)^{cot^3(x) cdot frac{1+sin(x)cos(beta x)}{sin(x)cos(alpha x)-sin(x)cos(beta x)} cdot frac{sin(x)cos(alpha x)-sin(x)cos(beta x)}{1+sin(x)cos(beta x)}} bigg) $$
$$e^{lim_{xto 0} big( frac{cos^3(x)sin(x)(cos(alpha x) - cos(beta x))}{sin(x)sin^2(x) (1+sin(x)cos(beta x))} big)}$$
Currently disregarding the exponent, because the font is small.
$$lim_{xto 0} big( frac{cos(alpha x) - cos(beta x)}{sin^2(x)} big)$$
Through Taylor's expansion,
$$lim_{xto 0} big( frac{1-frac{alpha^2x^2}{2} -1 + frac{beta^2x^2}{2}}{x^2} big)$$
And the answer is
$$e^{frac{beta^2 - alpha^2}{2}}.$$
The answer in the book, however, is $e^{beta^2 - alpha^2}$. The formula for the difference of two cosines also leaves me with $e^{frac{beta^2 - alpha^2}{2}}$, and I feel like there's a typo, but maybe I went wrong somewhere?
calculus limits proof-verification
calculus limits proof-verification
edited Nov 18 at 9:22
Kemono Chen
1,709330
asked Nov 18 at 9:08
fragileradius
278114
edited Nov 18 at 9:22
Kemono Chen
1,709330
asked Nov 18 at 9:08
fragileradius
278114
edited Nov 18 at 9:22
Kemono Chen
1,709330
edited Nov 18 at 9:22
Kemono Chen
1,709330
edited Nov 18 at 9:22
Kemono Chen
1,709330
1,709330
asked Nov 18 at 9:08
fragileradius
278114
asked Nov 18 at 9:08
fragileradius
278114
asked Nov 18 at 9:08
fragileradius
278114
278114
2
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26
add a comment |
2
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26
2
2
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
There's a typo, as affirmed in the comments.
answered Nov 18 at 9:28
fragileradius
278114
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There's a typo, as affirmed in the comments.
answered Nov 18 at 9:28
fragileradius
278114
add a comment |
up vote
1
down vote
accepted
There's a typo, as affirmed in the comments.
answered Nov 18 at 9:28
fragileradius
278114
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There's a typo, as affirmed in the comments.
answered Nov 18 at 9:28
fragileradius
278114
There's a typo, as affirmed in the comments.
answered Nov 18 at 9:28
fragileradius
278114
answered Nov 18 at 9:28
fragileradius
278114
answered Nov 18 at 9:28
fragileradius
278114
answered Nov 18 at 9:28
fragileradius
278114
278114
add a comment |
add a comment |
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2
Your book has a typo.
– Paramanand Singh
Nov 18 at 9:11
@ParamanandSingh Thanks! Do you suggest I delete this post, answer to myself and accept my own answer, or wait for another user to post the answer so I could accept it? I don't want my question hanging adding to the pile of unanswered ones.
– fragileradius
Nov 18 at 9:16
Answer yourself and accept it.
– Paramanand Singh
Nov 18 at 9:26