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Find the limit $lim_xto0fractan6x)sin3x$ without using L'hopital
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I'm trying to compute the following limit: $$lim_xto0fractan6xsin3x$$
I really have no idea how to start it. I tried rewriting $tan6x$ in terms of $sin6x$ and $cos6x$ but wasn't able to simplify the expression. How do I go about this?
limits limits-without-lhopital
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
asked Jul 20 at 6:10
Trey
292112
add a comment |Â
up vote
2
down vote
favorite
I'm trying to compute the following limit: $$lim_xto0fractan6xsin3x$$
I really have no idea how to start it. I tried rewriting $tan6x$ in terms of $sin6x$ and $cos6x$ but wasn't able to simplify the expression. How do I go about this?
limits limits-without-lhopital
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
asked Jul 20 at 6:10
Trey
292112
1
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to compute the following limit: $$lim_xto0fractan6xsin3x$$
I really have no idea how to start it. I tried rewriting $tan6x$ in terms of $sin6x$ and $cos6x$ but wasn't able to simplify the expression. How do I go about this?
limits limits-without-lhopital
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
asked Jul 20 at 6:10
Trey
292112
I'm trying to compute the following limit: $$lim_xto0fractan6xsin3x$$
I really have no idea how to start it. I tried rewriting $tan6x$ in terms of $sin6x$ and $cos6x$ but wasn't able to simplify the expression. How do I go about this?
limits limits-without-lhopital
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
asked Jul 20 at 6:10
Trey
292112
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
edited Jul 20 at 6:11
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 20 at 6:10
Trey
292112
asked Jul 20 at 6:10
Trey
292112
asked Jul 20 at 6:10
Trey
292112
292112
1
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25
add a comment |Â
1
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25
1
1
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
8
down vote
accepted
When $sin(3x)ne0$,
$$
beginalign
fractan(6x)sin(3x)
&=fracsin(6x)cos(6x)sin(3x)\
&=frac2sin(3x)cos(3x)cos(6x)sin(3x)\
&=2fraccos(3x)cos(6x)
endalign
$$
answered Jul 20 at 6:17
robjohn♦
258k26297612
add a comment |Â
up vote
5
down vote
By standard limit $fracsin ttto 1$ as $tto 0$, we have
$$fractan(6x)sin(3x)=fractan(6x)6xfrac3xsin(3x)frac6x3xto 1cdot 1cdot 2=2$$
answered Jul 20 at 6:12
gimusi
65.4k73584
add a comment |Â
up vote
4
down vote
With $tan6x=dfrac2tan3x1-tan^23x=dfrac2sin3xcos3x(1-tan^23x)$ then
$$lim_xto0dfractan6xsin3x=lim_xto0dfrac2cos3x(1-tan^23x)=2$$
answered Jul 20 at 6:15
Nosrati
19.5k41544
add a comment |Â
up vote
2
down vote
For fun:
Note: $tan 6x = dfrac2tan 3x1-tan^2 3x$.
$dfrac1cos 3 xdfrac tan 6xtan 3x=$
$dfrac1 cos 3x dfrac21-tan^2 3x.$
The limit is?
answered Jul 20 at 7:13
Peter Szilas
7,9352617
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
add a comment |Â
up vote
2
down vote
$$ lim _xto 0 frac tan 6xsin 3x \= lim _xto 0 frac sin 6xsin 3xcos 6x\=lim _xto 0 frac 1 cos 6x lim _xto 0 frac sin 6xsin3x\ =lim _xto 0 frac 2sin 3x cos 3xsin3x= lim _xto 0 2 cos 3x =2$$
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
When $sin(3x)ne0$,
$$
beginalign
fractan(6x)sin(3x)
&=fracsin(6x)cos(6x)sin(3x)\
&=frac2sin(3x)cos(3x)cos(6x)sin(3x)\
&=2fraccos(3x)cos(6x)
endalign
$$
answered Jul 20 at 6:17
robjohn♦
258k26297612
add a comment |Â
up vote
8
down vote
accepted
When $sin(3x)ne0$,
$$
beginalign
fractan(6x)sin(3x)
&=fracsin(6x)cos(6x)sin(3x)\
&=frac2sin(3x)cos(3x)cos(6x)sin(3x)\
&=2fraccos(3x)cos(6x)
endalign
$$
answered Jul 20 at 6:17
robjohn♦
258k26297612
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
When $sin(3x)ne0$,
$$
beginalign
fractan(6x)sin(3x)
&=fracsin(6x)cos(6x)sin(3x)\
&=frac2sin(3x)cos(3x)cos(6x)sin(3x)\
&=2fraccos(3x)cos(6x)
endalign
$$
answered Jul 20 at 6:17
robjohn♦
258k26297612
When $sin(3x)ne0$,
$$
beginalign
fractan(6x)sin(3x)
&=fracsin(6x)cos(6x)sin(3x)\
&=frac2sin(3x)cos(3x)cos(6x)sin(3x)\
&=2fraccos(3x)cos(6x)
endalign
$$
answered Jul 20 at 6:17
robjohn♦
258k26297612
answered Jul 20 at 6:17
robjohn♦
258k26297612
answered Jul 20 at 6:17
robjohn♦
258k26297612
answered Jul 20 at 6:17
robjohn♦
258k26297612
258k26297612
add a comment |Â
add a comment |Â
up vote
5
down vote
By standard limit $fracsin ttto 1$ as $tto 0$, we have
$$fractan(6x)sin(3x)=fractan(6x)6xfrac3xsin(3x)frac6x3xto 1cdot 1cdot 2=2$$
answered Jul 20 at 6:12
gimusi
65.4k73584
add a comment |Â
up vote
5
down vote
By standard limit $fracsin ttto 1$ as $tto 0$, we have
$$fractan(6x)sin(3x)=fractan(6x)6xfrac3xsin(3x)frac6x3xto 1cdot 1cdot 2=2$$
answered Jul 20 at 6:12
gimusi
65.4k73584
add a comment |Â
up vote
5
down vote
up vote
5
down vote
By standard limit $fracsin ttto 1$ as $tto 0$, we have
$$fractan(6x)sin(3x)=fractan(6x)6xfrac3xsin(3x)frac6x3xto 1cdot 1cdot 2=2$$
answered Jul 20 at 6:12
gimusi
65.4k73584
By standard limit $fracsin ttto 1$ as $tto 0$, we have
$$fractan(6x)sin(3x)=fractan(6x)6xfrac3xsin(3x)frac6x3xto 1cdot 1cdot 2=2$$
answered Jul 20 at 6:12
gimusi
65.4k73584
answered Jul 20 at 6:12
gimusi
65.4k73584
answered Jul 20 at 6:12
gimusi
65.4k73584
answered Jul 20 at 6:12
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
up vote
4
down vote
With $tan6x=dfrac2tan3x1-tan^23x=dfrac2sin3xcos3x(1-tan^23x)$ then
$$lim_xto0dfractan6xsin3x=lim_xto0dfrac2cos3x(1-tan^23x)=2$$
answered Jul 20 at 6:15
Nosrati
19.5k41544
add a comment |Â
up vote
4
down vote
With $tan6x=dfrac2tan3x1-tan^23x=dfrac2sin3xcos3x(1-tan^23x)$ then
$$lim_xto0dfractan6xsin3x=lim_xto0dfrac2cos3x(1-tan^23x)=2$$
answered Jul 20 at 6:15
Nosrati
19.5k41544
add a comment |Â
up vote
4
down vote
up vote
4
down vote
With $tan6x=dfrac2tan3x1-tan^23x=dfrac2sin3xcos3x(1-tan^23x)$ then
$$lim_xto0dfractan6xsin3x=lim_xto0dfrac2cos3x(1-tan^23x)=2$$
answered Jul 20 at 6:15
Nosrati
19.5k41544
With $tan6x=dfrac2tan3x1-tan^23x=dfrac2sin3xcos3x(1-tan^23x)$ then
$$lim_xto0dfractan6xsin3x=lim_xto0dfrac2cos3x(1-tan^23x)=2$$
answered Jul 20 at 6:15
Nosrati
19.5k41544
answered Jul 20 at 6:15
Nosrati
19.5k41544
answered Jul 20 at 6:15
Nosrati
19.5k41544
answered Jul 20 at 6:15
Nosrati
19.5k41544
19.5k41544
add a comment |Â
add a comment |Â
up vote
2
down vote
For fun:
Note: $tan 6x = dfrac2tan 3x1-tan^2 3x$.
$dfrac1cos 3 xdfrac tan 6xtan 3x=$
$dfrac1 cos 3x dfrac21-tan^2 3x.$
The limit is?
answered Jul 20 at 7:13
Peter Szilas
7,9352617
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
add a comment |Â
up vote
2
down vote
For fun:
Note: $tan 6x = dfrac2tan 3x1-tan^2 3x$.
$dfrac1cos 3 xdfrac tan 6xtan 3x=$
$dfrac1 cos 3x dfrac21-tan^2 3x.$
The limit is?
answered Jul 20 at 7:13
Peter Szilas
7,9352617
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For fun:
Note: $tan 6x = dfrac2tan 3x1-tan^2 3x$.
$dfrac1cos 3 xdfrac tan 6xtan 3x=$
$dfrac1 cos 3x dfrac21-tan^2 3x.$
The limit is?
answered Jul 20 at 7:13
Peter Szilas
7,9352617
For fun:
Note: $tan 6x = dfrac2tan 3x1-tan^2 3x$.
$dfrac1cos 3 xdfrac tan 6xtan 3x=$
$dfrac1 cos 3x dfrac21-tan^2 3x.$
The limit is?
answered Jul 20 at 7:13
Peter Szilas
7,9352617
answered Jul 20 at 7:13
Peter Szilas
7,9352617
answered Jul 20 at 7:13
Peter Szilas
7,9352617
answered Jul 20 at 7:13
Peter Szilas
7,9352617
7,9352617
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
add a comment |Â
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
@user108128 I know Peter very well, his solution is very similar to your but I don’t think it is intentional. The main idea is the same but the algebraic manipulation is slightly different
– gimusi
Jul 20 at 7:29
2
2
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
@user108128 I think you misunderstood what Peter meant
– Jam
Jul 20 at 8:14
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
Gimusi.Thanks for your comment.User 108128's solution is quite similar. Have not looked at the answers thoroughly . Kind words from you:)))Greetings.
– Peter Szilas
Jul 20 at 8:19
add a comment |Â
up vote
2
down vote
$$ lim _xto 0 frac tan 6xsin 3x \= lim _xto 0 frac sin 6xsin 3xcos 6x\=lim _xto 0 frac 1 cos 6x lim _xto 0 frac sin 6xsin3x\ =lim _xto 0 frac 2sin 3x cos 3xsin3x= lim _xto 0 2 cos 3x =2$$
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
add a comment |Â
up vote
2
down vote
$$ lim _xto 0 frac tan 6xsin 3x \= lim _xto 0 frac sin 6xsin 3xcos 6x\=lim _xto 0 frac 1 cos 6x lim _xto 0 frac sin 6xsin3x\ =lim _xto 0 frac 2sin 3x cos 3xsin3x= lim _xto 0 2 cos 3x =2$$
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$ lim _xto 0 frac tan 6xsin 3x \= lim _xto 0 frac sin 6xsin 3xcos 6x\=lim _xto 0 frac 1 cos 6x lim _xto 0 frac sin 6xsin3x\ =lim _xto 0 frac 2sin 3x cos 3xsin3x= lim _xto 0 2 cos 3x =2$$
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
$$ lim _xto 0 frac tan 6xsin 3x \= lim _xto 0 frac sin 6xsin 3xcos 6x\=lim _xto 0 frac 1 cos 6x lim _xto 0 frac sin 6xsin3x\ =lim _xto 0 frac 2sin 3x cos 3xsin3x= lim _xto 0 2 cos 3x =2$$
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
edited Jul 20 at 8:05
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 7:52
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
add a comment |Â
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
@gimusi Thanks for the comment. I edited my solution.
– Mohammad Riazi-Kermani
Jul 20 at 8:05
add a comment |Â
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1
Use $sin 6x=2sin 3xcos 3x$.
– Lord Shark the Unknown
Jul 20 at 6:14
In this particular case trigonometric manipulation is fine and maybe the simpler method. For more general cases keep in mind the use of standard limits, as for example $$fractan (pi x)sin (ex)$$ which is not solveble by simply trigonometric manipulations.
– gimusi
Jul 20 at 7:25