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Evaluation of $lim_x to 0 fracsin(x+a) -sin(a)sin2x$.
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I am having trouble proving out the below:
$$lim_x to 0 fracsin(x+a) - sin(a)sin2x$$
I have got as far as below using $sin(a) + sin(b)$ in numerator and $sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$lim_x to 0fracsin(x)cos(a) + sin(a)[cos(x) - 1]2sin(x)cos(x)$$
I need to simplify it down to apply $displaystylelim_x to 0fracsin(x)x$.
calculus limits trigonometry
edited Aug 1 at 2:23
user 108128
18.9k41544
asked Jul 31 at 20:58
number8
7717
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up vote
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I am having trouble proving out the below:
$$lim_x to 0 fracsin(x+a) - sin(a)sin2x$$
I have got as far as below using $sin(a) + sin(b)$ in numerator and $sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$lim_x to 0fracsin(x)cos(a) + sin(a)[cos(x) - 1]2sin(x)cos(x)$$
I need to simplify it down to apply $displaystylelim_x to 0fracsin(x)x$.
calculus limits trigonometry
edited Aug 1 at 2:23
user 108128
18.9k41544
asked Jul 31 at 20:58
number8
7717
Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having trouble proving out the below:
$$lim_x to 0 fracsin(x+a) - sin(a)sin2x$$
I have got as far as below using $sin(a) + sin(b)$ in numerator and $sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$lim_x to 0fracsin(x)cos(a) + sin(a)[cos(x) - 1]2sin(x)cos(x)$$
I need to simplify it down to apply $displaystylelim_x to 0fracsin(x)x$.
calculus limits trigonometry
edited Aug 1 at 2:23
user 108128
18.9k41544
asked Jul 31 at 20:58
number8
7717
I am having trouble proving out the below:
$$lim_x to 0 fracsin(x+a) - sin(a)sin2x$$
I have got as far as below using $sin(a) + sin(b)$ in numerator and $sin(2a)$ in the denominator but am not sure how to expand further, or if these are the right rules to apply.
$$lim_x to 0fracsin(x)cos(a) + sin(a)[cos(x) - 1]2sin(x)cos(x)$$
I need to simplify it down to apply $displaystylelim_x to 0fracsin(x)x$.
calculus limits trigonometry
edited Aug 1 at 2:23
user 108128
18.9k41544
asked Jul 31 at 20:58
number8
7717
edited Aug 1 at 2:23
user 108128
18.9k41544
edited Aug 1 at 2:23
user 108128
18.9k41544
edited Aug 1 at 2:23
user 108128
18.9k41544
18.9k41544
asked Jul 31 at 20:58
number8
7717
asked Jul 31 at 20:58
number8
7717
asked Jul 31 at 20:58
number8
7717
7717
Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02
add a comment |Â
Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02
Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02
add a comment |Â
7 Answers
7
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up vote
1
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accepted
Continue with this:
$$fracsin x cos a2sin xcos x + fracsin a[cos x - 1]2sin xcos x$$
$$fraccos a2cos x - sin afrac2sin^2fracx24sinfracx2cosfracx2cos x$$
$$fraccos a2cos x - sin afracsinfracx22cosfracx2cos x$$
then
$$lim_xto0fraccos a2cos x - sin afracsinfracx22cosfracx2cos x=fraccos a2times1-0=fraccos a2$$
answered Jul 31 at 21:07
user 108128
18.9k41544
add a comment |Â
up vote
5
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$frac dda sin a = lim_limitsxto 0 frac sin (x+a) - sin ax = cos a\
lim_limitsxto 0 frac sin (x+a) - sin asin 2x = lim_limitsxto 0 frac (sin (x+a) - sin a)xxsin 2x = lim_limitsxto 0 (frac sin (x+a) - sin ax)(frac xsin 2x) = frac 12 cos a $
answered Jul 31 at 21:07
Doug M
39k31749
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
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up vote
2
down vote
Hint: use that $$sin(x)-sin(y)=2cosleft(fracx+y2right)sinleft(fracx-y2right)$$
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
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up vote
2
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Hint:
Rewrite the quotient as
$$fracsin(x+a)-sin ax ,fracxsin 2x=fracsin(x+a)-sin ax ,frac2xsin 2xfrac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
answered Jul 31 at 21:09
Bernard
110k635102
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Write the limit in the following fashion namely
$$
lim_x to 0 fracsin(x+a) - sin(a)sin2x=fraclim_xto 0fracsin(a+x)-sin(a)x2lim_xto 0fracsin 2x2x=fraccos a2
$$
answered Jul 31 at 21:09
Foobaz John
18k41245
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up vote
1
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You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$lim_xto 0fraccos x - 1x = lim_xto 0frac(cos x-1)(cos x + 1)x(cos x+1) = - lim_xto 0fracsin^2 xxcdotfrac 1cos x+1.$$ Can you finish from this?
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
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up vote
1
down vote
Use the sum to product formula
$$sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$beginalign
sin(x+a)-sin(a)~&=~2cdot cosleft(frac(x+a)+a2right)sinleft(frac(x+a)-a2right)\
&=~2cdotcosleft(fracx2+aright)sinleft(fracx2right)
endalign$$
Subsitute this into the limit yields to
$$beginalign
lim_xto 0frac2cdotcosleft(fracx2+aright)sinleft(fracx2right)2cdotsin(x)cos(x)~&=~lim_xto 0fraccosleft(fracx2+aright)sinleft(fracx2right)sin(x)cos(x)\
&=~lim_xto 0fraccosleft(fracx2+aright)cos(x)cdotlim_xto 0fracsinleft(fracx2right)x\
&=~cos(a)~cdotlim_hto 0fracsinleft(hright)2h\
&=~cos(a)~cdotfrac12lim_hto 0fracsinleft(hright)h\
&=~fraccos(a)2
endalign$$
Where in the end $fracx2$ was substituted by $h$.
answered Jul 31 at 21:11
mrtaurho
660117
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Continue with this:
$$fracsin x cos a2sin xcos x + fracsin a[cos x - 1]2sin xcos x$$
$$fraccos a2cos x - sin afrac2sin^2fracx24sinfracx2cosfracx2cos x$$
$$fraccos a2cos x - sin afracsinfracx22cosfracx2cos x$$
then
$$lim_xto0fraccos a2cos x - sin afracsinfracx22cosfracx2cos x=fraccos a2times1-0=fraccos a2$$
answered Jul 31 at 21:07
user 108128
18.9k41544
add a comment |Â
up vote
1
down vote
accepted
Continue with this:
$$fracsin x cos a2sin xcos x + fracsin a[cos x - 1]2sin xcos x$$
$$fraccos a2cos x - sin afrac2sin^2fracx24sinfracx2cosfracx2cos x$$
$$fraccos a2cos x - sin afracsinfracx22cosfracx2cos x$$
then
$$lim_xto0fraccos a2cos x - sin afracsinfracx22cosfracx2cos x=fraccos a2times1-0=fraccos a2$$
answered Jul 31 at 21:07
user 108128
18.9k41544
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Continue with this:
$$fracsin x cos a2sin xcos x + fracsin a[cos x - 1]2sin xcos x$$
$$fraccos a2cos x - sin afrac2sin^2fracx24sinfracx2cosfracx2cos x$$
$$fraccos a2cos x - sin afracsinfracx22cosfracx2cos x$$
then
$$lim_xto0fraccos a2cos x - sin afracsinfracx22cosfracx2cos x=fraccos a2times1-0=fraccos a2$$
answered Jul 31 at 21:07
user 108128
18.9k41544
Continue with this:
$$fracsin x cos a2sin xcos x + fracsin a[cos x - 1]2sin xcos x$$
$$fraccos a2cos x - sin afrac2sin^2fracx24sinfracx2cosfracx2cos x$$
$$fraccos a2cos x - sin afracsinfracx22cosfracx2cos x$$
then
$$lim_xto0fraccos a2cos x - sin afracsinfracx22cosfracx2cos x=fraccos a2times1-0=fraccos a2$$
answered Jul 31 at 21:07
user 108128
18.9k41544
answered Jul 31 at 21:07
user 108128
18.9k41544
answered Jul 31 at 21:07
user 108128
18.9k41544
answered Jul 31 at 21:07
user 108128
18.9k41544
18.9k41544
add a comment |Â
add a comment |Â
up vote
5
down vote
$frac dda sin a = lim_limitsxto 0 frac sin (x+a) - sin ax = cos a\
lim_limitsxto 0 frac sin (x+a) - sin asin 2x = lim_limitsxto 0 frac (sin (x+a) - sin a)xxsin 2x = lim_limitsxto 0 (frac sin (x+a) - sin ax)(frac xsin 2x) = frac 12 cos a $
answered Jul 31 at 21:07
Doug M
39k31749
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
add a comment |Â
up vote
5
down vote
$frac dda sin a = lim_limitsxto 0 frac sin (x+a) - sin ax = cos a\
lim_limitsxto 0 frac sin (x+a) - sin asin 2x = lim_limitsxto 0 frac (sin (x+a) - sin a)xxsin 2x = lim_limitsxto 0 (frac sin (x+a) - sin ax)(frac xsin 2x) = frac 12 cos a $
answered Jul 31 at 21:07
Doug M
39k31749
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$frac dda sin a = lim_limitsxto 0 frac sin (x+a) - sin ax = cos a\
lim_limitsxto 0 frac sin (x+a) - sin asin 2x = lim_limitsxto 0 frac (sin (x+a) - sin a)xxsin 2x = lim_limitsxto 0 (frac sin (x+a) - sin ax)(frac xsin 2x) = frac 12 cos a $
answered Jul 31 at 21:07
Doug M
39k31749
$frac dda sin a = lim_limitsxto 0 frac sin (x+a) - sin ax = cos a\
lim_limitsxto 0 frac sin (x+a) - sin asin 2x = lim_limitsxto 0 frac (sin (x+a) - sin a)xxsin 2x = lim_limitsxto 0 (frac sin (x+a) - sin ax)(frac xsin 2x) = frac 12 cos a $
answered Jul 31 at 21:07
Doug M
39k31749
answered Jul 31 at 21:07
Doug M
39k31749
answered Jul 31 at 21:07
Doug M
39k31749
answered Jul 31 at 21:07
Doug M
39k31749
39k31749
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
add a comment |Â
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
2
2
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
First thing I thought of as well. I love approaches that use the definition of a derivative.
– Sorfosh
Jul 31 at 21:08
add a comment |Â
up vote
2
down vote
Hint: use that $$sin(x)-sin(y)=2cosleft(fracx+y2right)sinleft(fracx-y2right)$$
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
add a comment |Â
up vote
2
down vote
Hint: use that $$sin(x)-sin(y)=2cosleft(fracx+y2right)sinleft(fracx-y2right)$$
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: use that $$sin(x)-sin(y)=2cosleft(fracx+y2right)sinleft(fracx-y2right)$$
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
Hint: use that $$sin(x)-sin(y)=2cosleft(fracx+y2right)sinleft(fracx-y2right)$$
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
answered Jul 31 at 21:01
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
add a comment |Â
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
– lab bhattacharjee
Aug 1 at 0:11
add a comment |Â
up vote
2
down vote
Hint:
Rewrite the quotient as
$$fracsin(x+a)-sin ax ,fracxsin 2x=fracsin(x+a)-sin ax ,frac2xsin 2xfrac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
answered Jul 31 at 21:09
Bernard
110k635102
add a comment |Â
up vote
2
down vote
Hint:
Rewrite the quotient as
$$fracsin(x+a)-sin ax ,fracxsin 2x=fracsin(x+a)-sin ax ,frac2xsin 2xfrac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
answered Jul 31 at 21:09
Bernard
110k635102
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
Rewrite the quotient as
$$fracsin(x+a)-sin ax ,fracxsin 2x=fracsin(x+a)-sin ax ,frac2xsin 2xfrac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
answered Jul 31 at 21:09
Bernard
110k635102
Hint:
Rewrite the quotient as
$$fracsin(x+a)-sin ax ,fracxsin 2x=fracsin(x+a)-sin ax ,frac2xsin 2xfrac 12,$$
and observe the first fraction is but a rate of variation from the value $a$.
answered Jul 31 at 21:09
Bernard
110k635102
answered Jul 31 at 21:09
Bernard
110k635102
answered Jul 31 at 21:09
Bernard
110k635102
answered Jul 31 at 21:09
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
up vote
2
down vote
Write the limit in the following fashion namely
$$
lim_x to 0 fracsin(x+a) - sin(a)sin2x=fraclim_xto 0fracsin(a+x)-sin(a)x2lim_xto 0fracsin 2x2x=fraccos a2
$$
answered Jul 31 at 21:09
Foobaz John
18k41245
add a comment |Â
up vote
2
down vote
Write the limit in the following fashion namely
$$
lim_x to 0 fracsin(x+a) - sin(a)sin2x=fraclim_xto 0fracsin(a+x)-sin(a)x2lim_xto 0fracsin 2x2x=fraccos a2
$$
answered Jul 31 at 21:09
Foobaz John
18k41245
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Write the limit in the following fashion namely
$$
lim_x to 0 fracsin(x+a) - sin(a)sin2x=fraclim_xto 0fracsin(a+x)-sin(a)x2lim_xto 0fracsin 2x2x=fraccos a2
$$
answered Jul 31 at 21:09
Foobaz John
18k41245
Write the limit in the following fashion namely
$$
lim_x to 0 fracsin(x+a) - sin(a)sin2x=fraclim_xto 0fracsin(a+x)-sin(a)x2lim_xto 0fracsin 2x2x=fraccos a2
$$
answered Jul 31 at 21:09
Foobaz John
18k41245
answered Jul 31 at 21:09
Foobaz John
18k41245
answered Jul 31 at 21:09
Foobaz John
18k41245
answered Jul 31 at 21:09
Foobaz John
18k41245
18k41245
add a comment |Â
add a comment |Â
up vote
1
down vote
You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$lim_xto 0fraccos x - 1x = lim_xto 0frac(cos x-1)(cos x + 1)x(cos x+1) = - lim_xto 0fracsin^2 xxcdotfrac 1cos x+1.$$ Can you finish from this?
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
add a comment |Â
up vote
1
down vote
You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$lim_xto 0fraccos x - 1x = lim_xto 0frac(cos x-1)(cos x + 1)x(cos x+1) = - lim_xto 0fracsin^2 xxcdotfrac 1cos x+1.$$ Can you finish from this?
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$lim_xto 0fraccos x - 1x = lim_xto 0frac(cos x-1)(cos x + 1)x(cos x+1) = - lim_xto 0fracsin^2 xxcdotfrac 1cos x+1.$$ Can you finish from this?
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
You're doing just fine with your approach. Break it into a sum of limits. The first term is easy. For the second, you need to know
$$lim_xto 0fraccos x - 1x = lim_xto 0frac(cos x-1)(cos x + 1)x(cos x+1) = - lim_xto 0fracsin^2 xxcdotfrac 1cos x+1.$$ Can you finish from this?
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
answered Jul 31 at 21:05
Ted Shifrin
59.4k44386
59.4k44386
add a comment |Â
add a comment |Â
up vote
1
down vote
Use the sum to product formula
$$sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$beginalign
sin(x+a)-sin(a)~&=~2cdot cosleft(frac(x+a)+a2right)sinleft(frac(x+a)-a2right)\
&=~2cdotcosleft(fracx2+aright)sinleft(fracx2right)
endalign$$
Subsitute this into the limit yields to
$$beginalign
lim_xto 0frac2cdotcosleft(fracx2+aright)sinleft(fracx2right)2cdotsin(x)cos(x)~&=~lim_xto 0fraccosleft(fracx2+aright)sinleft(fracx2right)sin(x)cos(x)\
&=~lim_xto 0fraccosleft(fracx2+aright)cos(x)cdotlim_xto 0fracsinleft(fracx2right)x\
&=~cos(a)~cdotlim_hto 0fracsinleft(hright)2h\
&=~cos(a)~cdotfrac12lim_hto 0fracsinleft(hright)h\
&=~fraccos(a)2
endalign$$
Where in the end $fracx2$ was substituted by $h$.
answered Jul 31 at 21:11
mrtaurho
660117
add a comment |Â
up vote
1
down vote
Use the sum to product formula
$$sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$beginalign
sin(x+a)-sin(a)~&=~2cdot cosleft(frac(x+a)+a2right)sinleft(frac(x+a)-a2right)\
&=~2cdotcosleft(fracx2+aright)sinleft(fracx2right)
endalign$$
Subsitute this into the limit yields to
$$beginalign
lim_xto 0frac2cdotcosleft(fracx2+aright)sinleft(fracx2right)2cdotsin(x)cos(x)~&=~lim_xto 0fraccosleft(fracx2+aright)sinleft(fracx2right)sin(x)cos(x)\
&=~lim_xto 0fraccosleft(fracx2+aright)cos(x)cdotlim_xto 0fracsinleft(fracx2right)x\
&=~cos(a)~cdotlim_hto 0fracsinleft(hright)2h\
&=~cos(a)~cdotfrac12lim_hto 0fracsinleft(hright)h\
&=~fraccos(a)2
endalign$$
Where in the end $fracx2$ was substituted by $h$.
answered Jul 31 at 21:11
mrtaurho
660117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Use the sum to product formula
$$sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$beginalign
sin(x+a)-sin(a)~&=~2cdot cosleft(frac(x+a)+a2right)sinleft(frac(x+a)-a2right)\
&=~2cdotcosleft(fracx2+aright)sinleft(fracx2right)
endalign$$
Subsitute this into the limit yields to
$$beginalign
lim_xto 0frac2cdotcosleft(fracx2+aright)sinleft(fracx2right)2cdotsin(x)cos(x)~&=~lim_xto 0fraccosleft(fracx2+aright)sinleft(fracx2right)sin(x)cos(x)\
&=~lim_xto 0fraccosleft(fracx2+aright)cos(x)cdotlim_xto 0fracsinleft(fracx2right)x\
&=~cos(a)~cdotlim_hto 0fracsinleft(hright)2h\
&=~cos(a)~cdotfrac12lim_hto 0fracsinleft(hright)h\
&=~fraccos(a)2
endalign$$
Where in the end $fracx2$ was substituted by $h$.
answered Jul 31 at 21:11
mrtaurho
660117
Use the sum to product formula
$$sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$$
Plugging in $x=x+a$ and $y=a$ leads to
$$beginalign
sin(x+a)-sin(a)~&=~2cdot cosleft(frac(x+a)+a2right)sinleft(frac(x+a)-a2right)\
&=~2cdotcosleft(fracx2+aright)sinleft(fracx2right)
endalign$$
Subsitute this into the limit yields to
$$beginalign
lim_xto 0frac2cdotcosleft(fracx2+aright)sinleft(fracx2right)2cdotsin(x)cos(x)~&=~lim_xto 0fraccosleft(fracx2+aright)sinleft(fracx2right)sin(x)cos(x)\
&=~lim_xto 0fraccosleft(fracx2+aright)cos(x)cdotlim_xto 0fracsinleft(fracx2right)x\
&=~cos(a)~cdotlim_hto 0fracsinleft(hright)2h\
&=~cos(a)~cdotfrac12lim_hto 0fracsinleft(hright)h\
&=~fraccos(a)2
endalign$$
Where in the end $fracx2$ was substituted by $h$.
answered Jul 31 at 21:11
mrtaurho
660117
edited Jul 31 at 21:16
answered Jul 31 at 21:11
mrtaurho
660117
answered Jul 31 at 21:11
mrtaurho
660117
answered Jul 31 at 21:11
mrtaurho
660117
660117
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Maybe try to use $sin(x)-sin(y) = 2cdot cosleft(fracx+y2right)sinleft(fracx-y2right)$
– mrtaurho
Jul 31 at 21:01
Are you allowed to use L'Hopital's rule?
– Mark
Jul 31 at 21:02