Evaluating the limit of $frac{lnsin mx}{ln sin x}$

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.

asked Nov 16 at 23:04

Lowie

142

New contributor

1

Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08

add a comment |

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.

asked Nov 16 at 23:04

Lowie

142

New contributor

1

Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08

add a comment |

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.

asked Nov 16 at 23:04

Lowie

142

New contributor

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

asked Nov 16 at 23:04

Lowie

142

New contributor

asked Nov 16 at 23:04

Lowie

142

New contributor

asked Nov 16 at 23:04

Lowie

142

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.

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

Immediate aplication of L'Hopital's Rule.
– Kavi Rama Murthy
Nov 16 at 23:08

add a comment |

3 Answers
3

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.

edited Nov 16 at 23:12

answered Nov 16 at 23:07

gimusi

86k74292

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
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 |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

Lowie is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3001747%2fevaluating-the-limit-of-frac-ln-sin-mx-ln-sin-x%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

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.

edited Nov 16 at 23:12

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.

edited Nov 16 at 23:12

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.

edited Nov 16 at 23:12

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.

edited Nov 16 at 23:12

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

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

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

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

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

add a comment |

Lowie is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Lowie is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vrftsjtryk