Evaluating the limit of $frac{lnsin mx}{ln sin x}$
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I'd need some help evaluating this limit:
$$lim_{x to 0} frac{lnsin mx}{ln sin x}$$
I know it's supposed to equal 1 but I'm not sure how to get there.
real-analysis limits
asked Nov 16 at 23:04
Lowie
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I'd need some help evaluating this limit:
$$lim_{x to 0} frac{lnsin mx}{ln sin x}$$
I know it's supposed to equal 1 but I'm not sure how to get there.
real-analysis limits
asked Nov 16 at 23:04
Lowie
142
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08
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up vote
0
down vote
favorite
I'd need some help evaluating this limit:
$$lim_{x to 0} frac{lnsin mx}{ln sin x}$$
I know it's supposed to equal 1 but I'm not sure how to get there.
real-analysis limits
asked Nov 16 at 23:04
Lowie
142
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'd need some help evaluating this limit:
$$lim_{x to 0} frac{lnsin mx}{ln sin x}$$
I know it's supposed to equal 1 but I'm not sure how to get there.
real-analysis limits
real-analysis limits
asked Nov 16 at 23:04
Lowie
142
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 23:04
Lowie
142
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 23:04
Lowie
142
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Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 16 at 23:04
Lowie
142
asked Nov 16 at 23:04
Lowie
142
142
New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Lowie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
1
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08
add a comment |
1
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08
1
1
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08
add a comment |
3 Answers
3
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2
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HINT
We need $x>0$ and $m>0$ then by standard limits
$$frac{lnsin mx}{ln sin x}=frac{lnfrac{sin mx}{mx}+ln (mx)}{ln frac{sin x}x+ln x}=frac{lnfrac{sin mx}{mx}+ln x+ln m}{ln frac{sin x}x+ln x}$$
then recall that
- $frac{sin mx}{mx} to 1 implies lnfrac{sin mx}{mx}to 0$
- $frac{sin x}{x} to 1 implies lnfrac{sin x}{x}to 0$
and then the limit is determined by the $ln x$ terms.
answered Nov 16 at 23:07
gimusi
86k74292
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up vote
2
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You want $1+lim_{xto 0}dfrac{lnfrac{sin mx}{sin x}}{lnsin x}$. The numerator $to ln m$, the denominator $to -infty$. The result is therefore $1+0=1$.
answered Nov 16 at 23:29
J.G.
18.4k21932
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According to L^Hopital's rule we have $$lim_{x to 0} frac{lnsin mx}{ln sin x}=lim_{x to 0} frac{mcos mx}{cos x}{sin xover sin mx}=1$$
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT
We need $x>0$ and $m>0$ then by standard limits
$$frac{lnsin mx}{ln sin x}=frac{lnfrac{sin mx}{mx}+ln (mx)}{ln frac{sin x}x+ln x}=frac{lnfrac{sin mx}{mx}+ln x+ln m}{ln frac{sin x}x+ln x}$$
then recall that
- $frac{sin mx}{mx} to 1 implies lnfrac{sin mx}{mx}to 0$
- $frac{sin x}{x} to 1 implies lnfrac{sin x}{x}to 0$
and then the limit is determined by the $ln x$ terms.
answered Nov 16 at 23:07
gimusi
86k74292
add a comment |
up vote
2
down vote
HINT
We need $x>0$ and $m>0$ then by standard limits
$$frac{lnsin mx}{ln sin x}=frac{lnfrac{sin mx}{mx}+ln (mx)}{ln frac{sin x}x+ln x}=frac{lnfrac{sin mx}{mx}+ln x+ln m}{ln frac{sin x}x+ln x}$$
then recall that
- $frac{sin mx}{mx} to 1 implies lnfrac{sin mx}{mx}to 0$
- $frac{sin x}{x} to 1 implies lnfrac{sin x}{x}to 0$
and then the limit is determined by the $ln x$ terms.
answered Nov 16 at 23:07
gimusi
86k74292
add a comment |
up vote
2
down vote
up vote
2
down vote
HINT
We need $x>0$ and $m>0$ then by standard limits
$$frac{lnsin mx}{ln sin x}=frac{lnfrac{sin mx}{mx}+ln (mx)}{ln frac{sin x}x+ln x}=frac{lnfrac{sin mx}{mx}+ln x+ln m}{ln frac{sin x}x+ln x}$$
then recall that
- $frac{sin mx}{mx} to 1 implies lnfrac{sin mx}{mx}to 0$
- $frac{sin x}{x} to 1 implies lnfrac{sin x}{x}to 0$
and then the limit is determined by the $ln x$ terms.
answered Nov 16 at 23:07
gimusi
86k74292
HINT
We need $x>0$ and $m>0$ then by standard limits
$$frac{lnsin mx}{ln sin x}=frac{lnfrac{sin mx}{mx}+ln (mx)}{ln frac{sin x}x+ln x}=frac{lnfrac{sin mx}{mx}+ln x+ln m}{ln frac{sin x}x+ln x}$$
then recall that
- $frac{sin mx}{mx} to 1 implies lnfrac{sin mx}{mx}to 0$
- $frac{sin x}{x} to 1 implies lnfrac{sin x}{x}to 0$
and then the limit is determined by the $ln x$ terms.
answered Nov 16 at 23:07
gimusi
86k74292
edited Nov 16 at 23:12
answered Nov 16 at 23:07
gimusi
86k74292
answered Nov 16 at 23:07
gimusi
86k74292
answered Nov 16 at 23:07
gimusi
86k74292
86k74292
add a comment |
add a comment |
up vote
2
down vote
You want $1+lim_{xto 0}dfrac{lnfrac{sin mx}{sin x}}{lnsin x}$. The numerator $to ln m$, the denominator $to -infty$. The result is therefore $1+0=1$.
answered Nov 16 at 23:29
J.G.
18.4k21932
add a comment |
up vote
2
down vote
You want $1+lim_{xto 0}dfrac{lnfrac{sin mx}{sin x}}{lnsin x}$. The numerator $to ln m$, the denominator $to -infty$. The result is therefore $1+0=1$.
answered Nov 16 at 23:29
J.G.
18.4k21932
add a comment |
up vote
2
down vote
up vote
2
down vote
You want $1+lim_{xto 0}dfrac{lnfrac{sin mx}{sin x}}{lnsin x}$. The numerator $to ln m$, the denominator $to -infty$. The result is therefore $1+0=1$.
answered Nov 16 at 23:29
J.G.
18.4k21932
You want $1+lim_{xto 0}dfrac{lnfrac{sin mx}{sin x}}{lnsin x}$. The numerator $to ln m$, the denominator $to -infty$. The result is therefore $1+0=1$.
answered Nov 16 at 23:29
J.G.
18.4k21932
answered Nov 16 at 23:29
J.G.
18.4k21932
answered Nov 16 at 23:29
J.G.
18.4k21932
answered Nov 16 at 23:29
J.G.
18.4k21932
18.4k21932
add a comment |
add a comment |
up vote
0
down vote
According to L^Hopital's rule we have $$lim_{x to 0} frac{lnsin mx}{ln sin x}=lim_{x to 0} frac{mcos mx}{cos x}{sin xover sin mx}=1$$
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
add a comment |
up vote
0
down vote
According to L^Hopital's rule we have $$lim_{x to 0} frac{lnsin mx}{ln sin x}=lim_{x to 0} frac{mcos mx}{cos x}{sin xover sin mx}=1$$
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
add a comment |
up vote
0
down vote
up vote
0
down vote
According to L^Hopital's rule we have $$lim_{x to 0} frac{lnsin mx}{ln sin x}=lim_{x to 0} frac{mcos mx}{cos x}{sin xover sin mx}=1$$
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
According to L^Hopital's rule we have $$lim_{x to 0} frac{lnsin mx}{ln sin x}=lim_{x to 0} frac{mcos mx}{cos x}{sin xover sin mx}=1$$
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
answered Nov 16 at 23:08
Mostafa Ayaz
11.5k3733
11.5k3733
add a comment |
add a comment |
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1
Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08