find the limit of $lim limits_ xto 0 frac sin ^ -1 x sin3x $ without using l'Hopital's rule
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I'm new to calculus and i'm not sure how to deal with the inverse sin here
$$lim limits_ xto 0 leftfrac sin ^ -1 x sin3x right$$
limits limits-without-lhopital
edited Aug 16 at 9:35
mathreadler
13.7k71957
asked Aug 16 at 9:17
S113
143
add a comment |Â
up vote
1
down vote
favorite
I'm new to calculus and i'm not sure how to deal with the inverse sin here
$$lim limits_ xto 0 leftfrac sin ^ -1 x sin3x right$$
limits limits-without-lhopital
edited Aug 16 at 9:35
mathreadler
13.7k71957
asked Aug 16 at 9:17
S113
143
Equivalent infinitesimals.
– xbh
Aug 16 at 9:20
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm new to calculus and i'm not sure how to deal with the inverse sin here
$$lim limits_ xto 0 leftfrac sin ^ -1 x sin3x right$$
limits limits-without-lhopital
edited Aug 16 at 9:35
mathreadler
13.7k71957
asked Aug 16 at 9:17
S113
143
I'm new to calculus and i'm not sure how to deal with the inverse sin here
$$lim limits_ xto 0 leftfrac sin ^ -1 x sin3x right$$
limits limits-without-lhopital
edited Aug 16 at 9:35
mathreadler
13.7k71957
asked Aug 16 at 9:17
S113
143
edited Aug 16 at 9:35
mathreadler
13.7k71957
edited Aug 16 at 9:35
mathreadler
13.7k71957
edited Aug 16 at 9:35
mathreadler
13.7k71957
13.7k71957
asked Aug 16 at 9:17
S113
143
asked Aug 16 at 9:17
S113
143
asked Aug 16 at 9:17
S113
143
143
Equivalent infinitesimals.
– xbh
Aug 16 at 9:20
add a comment |Â
Equivalent infinitesimals.
– xbh
Aug 16 at 9:20
Equivalent infinitesimals.
– xbh
Aug 16 at 9:20
Equivalent infinitesimals.
– xbh
Aug 16 at 9:20
add a comment |Â
4 Answers
4
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oldest
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up vote
4
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accepted
Hint
$$lim_xto 0 dfracarcsin xx=lim_xto 0 dfracxsin x=1.$$
answered Aug 16 at 9:23
mfl
24.7k12040
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up vote
3
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$$lim limits_ xto 0 frac sin ^ -1 (x) sin(3x) =\$$
We substitute $x$ by $sin (t)$. This is possible because $sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get
$$=lim limits_ tto 0 frac sin ^ -1 (sin(t) )sin(3sin(t)) =\$$
$$=lim limits_ tto 0 frac tsin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)lim limits_ tto 0 frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ tto 0 frac 3sin(t)sin(3sin(t)) = $$
and if we substitute $3 sin t$ by $u$ then we get
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ uto 0 frac usin(u) = $$
$$=frac13$$
because $$lim limits_ yto 0 frac ysin(y)=1$$
answered Aug 16 at 10:07
miracle173
7,16922247
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up vote
2
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With substitution $xtosin t$ you have
$$lim_xto 0fracsin^-1xsin3x=lim_tto 0fractsin(3sin t)=lim_tto 0frac3sin tsin(3sin t)cdotfract3sin t=colorbluedfrac13$$
answered Aug 16 at 9:29
Nosrati
20.6k41644
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
add a comment |Â
up vote
1
down vote
$sin ^-1x= t$
$limlimits_x→0sin 3t→ 3t$
Therefore:
$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t=dfrac13$
edited Aug 16 at 10:09
miracle173
7,16922247
answered Aug 16 at 9:43
sirous
876511
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint
$$lim_xto 0 dfracarcsin xx=lim_xto 0 dfracxsin x=1.$$
answered Aug 16 at 9:23
mfl
24.7k12040
add a comment |Â
up vote
4
down vote
accepted
Hint
$$lim_xto 0 dfracarcsin xx=lim_xto 0 dfracxsin x=1.$$
answered Aug 16 at 9:23
mfl
24.7k12040
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint
$$lim_xto 0 dfracarcsin xx=lim_xto 0 dfracxsin x=1.$$
answered Aug 16 at 9:23
mfl
24.7k12040
Hint
$$lim_xto 0 dfracarcsin xx=lim_xto 0 dfracxsin x=1.$$
answered Aug 16 at 9:23
mfl
24.7k12040
answered Aug 16 at 9:23
mfl
24.7k12040
answered Aug 16 at 9:23
mfl
24.7k12040
answered Aug 16 at 9:23
mfl
24.7k12040
24.7k12040
add a comment |Â
add a comment |Â
up vote
3
down vote
$$lim limits_ xto 0 frac sin ^ -1 (x) sin(3x) =\$$
We substitute $x$ by $sin (t)$. This is possible because $sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get
$$=lim limits_ tto 0 frac sin ^ -1 (sin(t) )sin(3sin(t)) =\$$
$$=lim limits_ tto 0 frac tsin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)lim limits_ tto 0 frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ tto 0 frac 3sin(t)sin(3sin(t)) = $$
and if we substitute $3 sin t$ by $u$ then we get
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ uto 0 frac usin(u) = $$
$$=frac13$$
because $$lim limits_ yto 0 frac ysin(y)=1$$
answered Aug 16 at 10:07
miracle173
7,16922247
add a comment |Â
up vote
3
down vote
$$lim limits_ xto 0 frac sin ^ -1 (x) sin(3x) =\$$
We substitute $x$ by $sin (t)$. This is possible because $sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get
$$=lim limits_ tto 0 frac sin ^ -1 (sin(t) )sin(3sin(t)) =\$$
$$=lim limits_ tto 0 frac tsin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)lim limits_ tto 0 frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ tto 0 frac 3sin(t)sin(3sin(t)) = $$
and if we substitute $3 sin t$ by $u$ then we get
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ uto 0 frac usin(u) = $$
$$=frac13$$
because $$lim limits_ yto 0 frac ysin(y)=1$$
answered Aug 16 at 10:07
miracle173
7,16922247
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$lim limits_ xto 0 frac sin ^ -1 (x) sin(3x) =\$$
We substitute $x$ by $sin (t)$. This is possible because $sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get
$$=lim limits_ tto 0 frac sin ^ -1 (sin(t) )sin(3sin(t)) =\$$
$$=lim limits_ tto 0 frac tsin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)lim limits_ tto 0 frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ tto 0 frac 3sin(t)sin(3sin(t)) = $$
and if we substitute $3 sin t$ by $u$ then we get
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ uto 0 frac usin(u) = $$
$$=frac13$$
because $$lim limits_ yto 0 frac ysin(y)=1$$
answered Aug 16 at 10:07
miracle173
7,16922247
$$lim limits_ xto 0 frac sin ^ -1 (x) sin(3x) =\$$
We substitute $x$ by $sin (t)$. This is possible because $sin$ is a strict increasing continous function near $0$ and $sin(0)=0$ and get
$$=lim limits_ tto 0 frac sin ^ -1 (sin(t) )sin(3sin(t)) =\$$
$$=lim limits_ tto 0 frac tsin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)lim limits_ tto 0 frac sin(t)sin(3sin(t)) = $$
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ tto 0 frac 3sin(t)sin(3sin(t)) = $$
and if we substitute $3 sin t$ by $u$ then we get
$$=lim limits_ tto 0 frac tsin(t)frac13lim limits_ uto 0 frac usin(u) = $$
$$=frac13$$
because $$lim limits_ yto 0 frac ysin(y)=1$$
answered Aug 16 at 10:07
miracle173
7,16922247
answered Aug 16 at 10:07
miracle173
7,16922247
answered Aug 16 at 10:07
miracle173
7,16922247
answered Aug 16 at 10:07
miracle173
7,16922247
7,16922247
add a comment |Â
add a comment |Â
up vote
2
down vote
With substitution $xtosin t$ you have
$$lim_xto 0fracsin^-1xsin3x=lim_tto 0fractsin(3sin t)=lim_tto 0frac3sin tsin(3sin t)cdotfract3sin t=colorbluedfrac13$$
answered Aug 16 at 9:29
Nosrati
20.6k41644
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
add a comment |Â
up vote
2
down vote
With substitution $xtosin t$ you have
$$lim_xto 0fracsin^-1xsin3x=lim_tto 0fractsin(3sin t)=lim_tto 0frac3sin tsin(3sin t)cdotfract3sin t=colorbluedfrac13$$
answered Aug 16 at 9:29
Nosrati
20.6k41644
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
add a comment |Â
up vote
2
down vote
up vote
2
down vote
With substitution $xtosin t$ you have
$$lim_xto 0fracsin^-1xsin3x=lim_tto 0fractsin(3sin t)=lim_tto 0frac3sin tsin(3sin t)cdotfract3sin t=colorbluedfrac13$$
answered Aug 16 at 9:29
Nosrati
20.6k41644
With substitution $xtosin t$ you have
$$lim_xto 0fracsin^-1xsin3x=lim_tto 0fractsin(3sin t)=lim_tto 0frac3sin tsin(3sin t)cdotfract3sin t=colorbluedfrac13$$
answered Aug 16 at 9:29
Nosrati
20.6k41644
edited Aug 16 at 9:44
answered Aug 16 at 9:29
Nosrati
20.6k41644
answered Aug 16 at 9:29
Nosrati
20.6k41644
answered Aug 16 at 9:29
Nosrati
20.6k41644
20.6k41644
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
add a comment |Â
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
parenthesis could be used to avoid confusion here.
– mathreadler
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
sure . . . . . .
– Nosrati
Aug 16 at 9:36
add a comment |Â
up vote
1
down vote
$sin ^-1x= t$
$limlimits_x→0sin 3t→ 3t$
Therefore:
$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t=dfrac13$
edited Aug 16 at 10:09
miracle173
7,16922247
answered Aug 16 at 9:43
sirous
876511
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
add a comment |Â
up vote
1
down vote
$sin ^-1x= t$
$limlimits_x→0sin 3t→ 3t$
Therefore:
$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t=dfrac13$
edited Aug 16 at 10:09
miracle173
7,16922247
answered Aug 16 at 9:43
sirous
876511
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$sin ^-1x= t$
$limlimits_x→0sin 3t→ 3t$
Therefore:
$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t=dfrac13$
edited Aug 16 at 10:09
miracle173
7,16922247
answered Aug 16 at 9:43
sirous
876511
$sin ^-1x= t$
$limlimits_x→0sin 3t→ 3t$
Therefore:
$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t=dfrac13$
edited Aug 16 at 10:09
miracle173
7,16922247
answered Aug 16 at 9:43
sirous
876511
edited Aug 16 at 10:09
miracle173
7,16922247
edited Aug 16 at 10:09
miracle173
7,16922247
edited Aug 16 at 10:09
miracle173
7,16922247
7,16922247
answered Aug 16 at 9:43
sirous
876511
answered Aug 16 at 9:43
sirous
876511
answered Aug 16 at 9:43
sirous
876511
876511
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
add a comment |Â
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
1
1
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
I think the notation and the format should be improved.
– miracle173
Aug 16 at 9:48
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
Indeed, $$limlimits_x→0sin 3t→ 3t$$ and $$limlimits_x→0 dfracsin^-1 xsin 3x=dfract3t$$ are not suitable to any decent answer.
– Did
Aug 16 at 11:12
add a comment |Â
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Equivalent infinitesimals.
– xbh
Aug 16 at 9:20