Mixing
Finding limit of $sin(x^2/2)/sqrt{2}sin^2(x/2)$ as $xrightarrow0$
$begingroup$
Can anybody help me find the limit as $x$ tends to $0$, for $$frac{sin(x^2/2)}{sqrt{2}sin^2(x/2)}.$$ How can I simplify the expression or use equivalent transformations to find a limit (without using L'Hospital)?
limits functions limits-without-lhopital
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
$endgroup$
add a comment |
$begingroup$
Can anybody help me find the limit as $x$ tends to $0$, for $$frac{sin(x^2/2)}{sqrt{2}sin^2(x/2)}.$$ How can I simplify the expression or use equivalent transformations to find a limit (without using L'Hospital)?
limits functions limits-without-lhopital
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
$endgroup$
2
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
4
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24
add a comment |
$begingroup$
Can anybody help me find the limit as $x$ tends to $0$, for $$frac{sin(x^2/2)}{sqrt{2}sin^2(x/2)}.$$ How can I simplify the expression or use equivalent transformations to find a limit (without using L'Hospital)?
limits functions limits-without-lhopital
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
$endgroup$
Can anybody help me find the limit as $x$ tends to $0$, for $$frac{sin(x^2/2)}{sqrt{2}sin^2(x/2)}.$$ How can I simplify the expression or use equivalent transformations to find a limit (without using L'Hospital)?
limits functions limits-without-lhopital
limits functions limits-without-lhopital
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
edited Jan 23 at 1:57
Alexander Gruber♦
20.2k25102173
20.2k25102173
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
asked Jan 22 at 15:16
Ieva BrakmaneIeva Brakmane
163
163
2
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
4
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24
add a comment |
2
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
4
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24
2
2
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
4
4
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I will be using just one fact to solve this, and that is,
$$lim_{xto 0} frac{ sin x}{x} = 1$$
So lets start!
$$lim_{xto 0} frac {sin frac {x^2}{2}}{sqrt{2} ; sin^2 frac {x}{2}}$$
Now, Multiply and divide by, $frac{x^2}{2} ;$ and $(frac {x}{2})^2$
The $frac{x^2}{2}$ term in the denominator along with $sin frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(frac {x}{2})^2$ in the numerator and $sin^2 frac {x}{2}$ in the denominator.
After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).
Resulting in the answer (cue the drumroll) $$sqrt 2$$
Edit: I have included the steps taken just as a measure
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
$endgroup$
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
|
show 1 more comment
$begingroup$
As $dfrac{sin x}x=1$, in multiplicative expressions you can replace $sin x$ by $x$.
Hence
$$lim_{xto0}frac{sinleft(dfrac{x^2}2right)}{sqrt2,sin^2left(dfrac x2right)}
=lim_{xto0}frac{dfrac{x^2}2}{sqrt2,left(dfrac x2right)^2}=sqrt2.$$
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I will be using just one fact to solve this, and that is,
$$lim_{xto 0} frac{ sin x}{x} = 1$$
So lets start!
$$lim_{xto 0} frac {sin frac {x^2}{2}}{sqrt{2} ; sin^2 frac {x}{2}}$$
Now, Multiply and divide by, $frac{x^2}{2} ;$ and $(frac {x}{2})^2$
The $frac{x^2}{2}$ term in the denominator along with $sin frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(frac {x}{2})^2$ in the numerator and $sin^2 frac {x}{2}$ in the denominator.
After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).
Resulting in the answer (cue the drumroll) $$sqrt 2$$
Edit: I have included the steps taken just as a measure
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
$endgroup$
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
|
show 1 more comment
$begingroup$
I will be using just one fact to solve this, and that is,
$$lim_{xto 0} frac{ sin x}{x} = 1$$
So lets start!
$$lim_{xto 0} frac {sin frac {x^2}{2}}{sqrt{2} ; sin^2 frac {x}{2}}$$
Now, Multiply and divide by, $frac{x^2}{2} ;$ and $(frac {x}{2})^2$
The $frac{x^2}{2}$ term in the denominator along with $sin frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(frac {x}{2})^2$ in the numerator and $sin^2 frac {x}{2}$ in the denominator.
After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).
Resulting in the answer (cue the drumroll) $$sqrt 2$$
Edit: I have included the steps taken just as a measure
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
$endgroup$
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
|
show 1 more comment
$begingroup$
I will be using just one fact to solve this, and that is,
$$lim_{xto 0} frac{ sin x}{x} = 1$$
So lets start!
$$lim_{xto 0} frac {sin frac {x^2}{2}}{sqrt{2} ; sin^2 frac {x}{2}}$$
Now, Multiply and divide by, $frac{x^2}{2} ;$ and $(frac {x}{2})^2$
The $frac{x^2}{2}$ term in the denominator along with $sin frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(frac {x}{2})^2$ in the numerator and $sin^2 frac {x}{2}$ in the denominator.
After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).
Resulting in the answer (cue the drumroll) $$sqrt 2$$
Edit: I have included the steps taken just as a measure
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
$endgroup$
I will be using just one fact to solve this, and that is,
$$lim_{xto 0} frac{ sin x}{x} = 1$$
So lets start!
$$lim_{xto 0} frac {sin frac {x^2}{2}}{sqrt{2} ; sin^2 frac {x}{2}}$$
Now, Multiply and divide by, $frac{x^2}{2} ;$ and $(frac {x}{2})^2$
The $frac{x^2}{2}$ term in the denominator along with $sin frac {x^2}{2}$ in the numerator tends to ONE. And so is the case with the $(frac {x}{2})^2$ in the numerator and $sin^2 frac {x}{2}$ in the denominator.
After that we see that the remaining $x^2$ terms cancel out (because we have one both in the numerator and the denominator).
Resulting in the answer (cue the drumroll) $$sqrt 2$$
Edit: I have included the steps taken just as a measure
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
edited Jan 22 at 18:27
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
answered Jan 22 at 15:46
The Jade EmperorThe Jade Emperor
908
908
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
|
show 1 more comment
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
The $sqrt 2$ which is in the denominator stays out of our calculation. After simplifying we get $$lim_{x to 0} frac{sin frac {x^2}{2}}{sin^2 frac{x}{2}} = 2$$.. and that $2$ cancels with $sqrt 2 $ to give the stated answer
$endgroup$
– The Jade Emperor
Jan 22 at 18:15
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. I will attach all detailed steps
$endgroup$
– The Jade Emperor
Jan 22 at 18:16
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
Ok. Thank you very much.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:25
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
yeah I've edited my answer to show the method. Hope you like it! If you still have questions feel free to ask! ((Otherwise upvote and accept the answer :p))
$endgroup$
– The Jade Emperor
Jan 22 at 18:28
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
$begingroup$
Thank you very much. Maybe you know any good source when one can get accustomized with these set of problems? I accepted the answer, but cannot upvote you as I don't have enough reputation points yet.
$endgroup$
– Ieva Brakmane
Jan 22 at 18:41
|
show 1 more comment
$begingroup$
As $dfrac{sin x}x=1$, in multiplicative expressions you can replace $sin x$ by $x$.
Hence
$$lim_{xto0}frac{sinleft(dfrac{x^2}2right)}{sqrt2,sin^2left(dfrac x2right)}
=lim_{xto0}frac{dfrac{x^2}2}{sqrt2,left(dfrac x2right)^2}=sqrt2.$$
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
As $dfrac{sin x}x=1$, in multiplicative expressions you can replace $sin x$ by $x$.
Hence
$$lim_{xto0}frac{sinleft(dfrac{x^2}2right)}{sqrt2,sin^2left(dfrac x2right)}
=lim_{xto0}frac{dfrac{x^2}2}{sqrt2,left(dfrac x2right)^2}=sqrt2.$$
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
As $dfrac{sin x}x=1$, in multiplicative expressions you can replace $sin x$ by $x$.
Hence
$$lim_{xto0}frac{sinleft(dfrac{x^2}2right)}{sqrt2,sin^2left(dfrac x2right)}
=lim_{xto0}frac{dfrac{x^2}2}{sqrt2,left(dfrac x2right)^2}=sqrt2.$$
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
$endgroup$
As $dfrac{sin x}x=1$, in multiplicative expressions you can replace $sin x$ by $x$.
Hence
$$lim_{xto0}frac{sinleft(dfrac{x^2}2right)}{sqrt2,sin^2left(dfrac x2right)}
=lim_{xto0}frac{dfrac{x^2}2}{sqrt2,left(dfrac x2right)^2}=sqrt2.$$
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
answered Jan 22 at 18:48
Yves DaoustYves Daoust
130k676227
130k676227
add a comment |
add a comment |
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2
$begingroup$
Please refer to this guide on how to typeset maths on this site. It's pretty hard to understand what is your expression you are working with
$endgroup$
– roman
Jan 22 at 15:21
$begingroup$
Use Taylor expansions.
$endgroup$
– Clayton
Jan 22 at 15:23
4
$begingroup$
HINT: Note that $$frac{sin(x^2/2)}{sqrt2 sin^2(x/2)}=left(frac{sin(x^2/2)}{x^2/2}right)left(frac{x^2}{2}right)frac{1}{sqrt2}left(frac{x/2}{sin(x/2)}right)^2left(frac{1}{(x/2)^2}right)$$Can you finish now?
$endgroup$
– Mark Viola
Jan 22 at 15:24