How to evaluate $lim_x to fracpi4 fracsqrt1-sqrtsin 2xpi-4x$
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I can not figure out why the limit does not exist ?
can anyone explain it ?
$$lim_x to fracpi4 fracsqrt1-sqrtsin 2xpi-4x$$
limits limits-without-lhopital
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
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up vote
0
down vote
favorite
I can not figure out why the limit does not exist ?
can anyone explain it ?
$$lim_x to fracpi4 fracsqrt1-sqrtsin 2xpi-4x$$
limits limits-without-lhopital
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I can not figure out why the limit does not exist ?
can anyone explain it ?
$$lim_x to fracpi4 fracsqrt1-sqrtsin 2xpi-4x$$
limits limits-without-lhopital
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
I can not figure out why the limit does not exist ?
can anyone explain it ?
$$lim_x to fracpi4 fracsqrt1-sqrtsin 2xpi-4x$$
limits limits-without-lhopital
limits limits-without-lhopital
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
edited Jan 25 '16 at 8:51
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
asked Jan 25 '16 at 8:50
Angelo Mark
3,98921239
3,98921239
There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24
add a comment |Â
There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24
There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24
add a comment |Â
2 Answers
2
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oldest
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up vote
2
down vote
accepted
$$1-sin2x=(cos x-sin x)^2$$
$$sqrt1-sin2x=|sin x-cos x|=sqrt2left|sinleft(x-dfracpi4right)right|$$
$=+(sin x-cos x)$ iff $xgedfracpi4$
What happens otherwise?
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
add a comment |Â
up vote
2
down vote
Considering $$A= fracsqrt1-sqrtsin 2xpi-4x$$ start making life easier setting $x=fracpi 4-y$. So $$A=fracsqrt1-sqrtcos (2 y)4 y$$ Now, use the fact that, for small $z$, $$cos(z)=1-fracz^22+fracz^424+Oleft(z^5right)$$ which makes $$cos(2y)=1-2 y^2+frac2 y^43+Oleft(y^5right)$$ Use the generalized binomial theorem to get $$sqrtcos (2 y)=1-y^2-fracy^46+Oleft(y^5right)$$ $$1-sqrtcos (2 y)=y^2+fracy^46+Oleft(y^5right)$$ $$sqrt1-sqrtcos (2 y)=|y| sqrt1+fracy^26+cdots$$ So, by the end $$A=frac 14 fracy y sqrt1+fracy^26+cdots$$ So, depending if $yto 0^+$, $Ato frac 14$ and if $yto 0^-$, $Ato -frac 14$.
So, the limit does not exist.
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$1-sin2x=(cos x-sin x)^2$$
$$sqrt1-sin2x=|sin x-cos x|=sqrt2left|sinleft(x-dfracpi4right)right|$$
$=+(sin x-cos x)$ iff $xgedfracpi4$
What happens otherwise?
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
add a comment |Â
up vote
2
down vote
accepted
$$1-sin2x=(cos x-sin x)^2$$
$$sqrt1-sin2x=|sin x-cos x|=sqrt2left|sinleft(x-dfracpi4right)right|$$
$=+(sin x-cos x)$ iff $xgedfracpi4$
What happens otherwise?
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$1-sin2x=(cos x-sin x)^2$$
$$sqrt1-sin2x=|sin x-cos x|=sqrt2left|sinleft(x-dfracpi4right)right|$$
$=+(sin x-cos x)$ iff $xgedfracpi4$
What happens otherwise?
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
$$1-sin2x=(cos x-sin x)^2$$
$$sqrt1-sin2x=|sin x-cos x|=sqrt2left|sinleft(x-dfracpi4right)right|$$
$=+(sin x-cos x)$ iff $xgedfracpi4$
What happens otherwise?
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
answered Jan 25 '16 at 8:56
lab bhattacharjee
216k14153265
216k14153265
add a comment |Â
add a comment |Â
up vote
2
down vote
Considering $$A= fracsqrt1-sqrtsin 2xpi-4x$$ start making life easier setting $x=fracpi 4-y$. So $$A=fracsqrt1-sqrtcos (2 y)4 y$$ Now, use the fact that, for small $z$, $$cos(z)=1-fracz^22+fracz^424+Oleft(z^5right)$$ which makes $$cos(2y)=1-2 y^2+frac2 y^43+Oleft(y^5right)$$ Use the generalized binomial theorem to get $$sqrtcos (2 y)=1-y^2-fracy^46+Oleft(y^5right)$$ $$1-sqrtcos (2 y)=y^2+fracy^46+Oleft(y^5right)$$ $$sqrt1-sqrtcos (2 y)=|y| sqrt1+fracy^26+cdots$$ So, by the end $$A=frac 14 fracy y sqrt1+fracy^26+cdots$$ So, depending if $yto 0^+$, $Ato frac 14$ and if $yto 0^-$, $Ato -frac 14$.
So, the limit does not exist.
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
add a comment |Â
up vote
2
down vote
Considering $$A= fracsqrt1-sqrtsin 2xpi-4x$$ start making life easier setting $x=fracpi 4-y$. So $$A=fracsqrt1-sqrtcos (2 y)4 y$$ Now, use the fact that, for small $z$, $$cos(z)=1-fracz^22+fracz^424+Oleft(z^5right)$$ which makes $$cos(2y)=1-2 y^2+frac2 y^43+Oleft(y^5right)$$ Use the generalized binomial theorem to get $$sqrtcos (2 y)=1-y^2-fracy^46+Oleft(y^5right)$$ $$1-sqrtcos (2 y)=y^2+fracy^46+Oleft(y^5right)$$ $$sqrt1-sqrtcos (2 y)=|y| sqrt1+fracy^26+cdots$$ So, by the end $$A=frac 14 fracy y sqrt1+fracy^26+cdots$$ So, depending if $yto 0^+$, $Ato frac 14$ and if $yto 0^-$, $Ato -frac 14$.
So, the limit does not exist.
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Considering $$A= fracsqrt1-sqrtsin 2xpi-4x$$ start making life easier setting $x=fracpi 4-y$. So $$A=fracsqrt1-sqrtcos (2 y)4 y$$ Now, use the fact that, for small $z$, $$cos(z)=1-fracz^22+fracz^424+Oleft(z^5right)$$ which makes $$cos(2y)=1-2 y^2+frac2 y^43+Oleft(y^5right)$$ Use the generalized binomial theorem to get $$sqrtcos (2 y)=1-y^2-fracy^46+Oleft(y^5right)$$ $$1-sqrtcos (2 y)=y^2+fracy^46+Oleft(y^5right)$$ $$sqrt1-sqrtcos (2 y)=|y| sqrt1+fracy^26+cdots$$ So, by the end $$A=frac 14 fracy y sqrt1+fracy^26+cdots$$ So, depending if $yto 0^+$, $Ato frac 14$ and if $yto 0^-$, $Ato -frac 14$.
So, the limit does not exist.
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
Considering $$A= fracsqrt1-sqrtsin 2xpi-4x$$ start making life easier setting $x=fracpi 4-y$. So $$A=fracsqrt1-sqrtcos (2 y)4 y$$ Now, use the fact that, for small $z$, $$cos(z)=1-fracz^22+fracz^424+Oleft(z^5right)$$ which makes $$cos(2y)=1-2 y^2+frac2 y^43+Oleft(y^5right)$$ Use the generalized binomial theorem to get $$sqrtcos (2 y)=1-y^2-fracy^46+Oleft(y^5right)$$ $$1-sqrtcos (2 y)=y^2+fracy^46+Oleft(y^5right)$$ $$sqrt1-sqrtcos (2 y)=|y| sqrt1+fracy^26+cdots$$ So, by the end $$A=frac 14 fracy y sqrt1+fracy^26+cdots$$ So, depending if $yto 0^+$, $Ato frac 14$ and if $yto 0^-$, $Ato -frac 14$.
So, the limit does not exist.
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
answered Jan 25 '16 at 9:31
Claude Leibovici
113k1155127
113k1155127
add a comment |Â
add a comment |Â
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There is something missing. What is the domain of study of your function?
– Martigan
Jan 25 '16 at 8:55
The limit doesn't exist because the value approached by the function differs depending on whether you get to $pi/4$ from the left or from the right. In particular, note that the denominator is negative for $x>pi/4$, making negative the whole function, whereas it's positive otherwise.
– Vincenzo Oliva
Jan 25 '16 at 9:26
Can you please tell me what book is that?
– Mylton Franklyn
Sep 4 at 5:24