Limit of goniometric function without l'Hospital's rule
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I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }$$
limits limits-without-lhopital
edited yesterday
gimusi
86k74292
asked yesterday
Lukáš Frajt
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Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
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I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }$$
limits limits-without-lhopital
edited yesterday
gimusi
86k74292
asked yesterday
Lukáš Frajt
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }$$
limits limits-without-lhopital
edited yesterday
gimusi
86k74292
asked yesterday
Lukáš Frajt
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm trying figure this out without l'Hospital's rule. But I don't know how should I start. Any hint, please?
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }$$
limits limits-without-lhopital
limits limits-without-lhopital
edited yesterday
gimusi
86k74292
asked yesterday
Lukáš Frajt
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
gimusi
86k74292
asked yesterday
Lukáš Frajt
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
gimusi
86k74292
edited yesterday
gimusi
86k74292
edited yesterday
gimusi
86k74292
86k74292
asked yesterday
Lukáš Frajt
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Lukáš Frajt
81
asked yesterday
Lukáš Frajt
81
81
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lukáš Frajt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday
add a comment |
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Set $t=frac pi2 - x,$
$$lim_{xto {piover 2}} frac {1-sin x}{(fracpi2 -x)^2}=lim_{tto {0}} frac {1-cos t}{t^2}=lim_{tto {0}} frac {2 sin ^2(t/2)}{4(t/2)^2}={1over 2}$$
answered yesterday
user376343
2,2331716
add a comment |
up vote
0
down vote
We have that
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }
=lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 } frac {1+sin x}{1+sin x}
=lim_{xto frac{pi}2} frac {1}{1+sin x}frac {sin^2left(fracpi2 -xright)}{left(fracpi2 -xright)^2 } =frac12$$
answered yesterday
gimusi
86k74292
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
Got it, thank you!
– Lukáš Frajt
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Set $t=frac pi2 - x,$
$$lim_{xto {piover 2}} frac {1-sin x}{(fracpi2 -x)^2}=lim_{tto {0}} frac {1-cos t}{t^2}=lim_{tto {0}} frac {2 sin ^2(t/2)}{4(t/2)^2}={1over 2}$$
answered yesterday
user376343
2,2331716
add a comment |
up vote
2
down vote
accepted
Set $t=frac pi2 - x,$
$$lim_{xto {piover 2}} frac {1-sin x}{(fracpi2 -x)^2}=lim_{tto {0}} frac {1-cos t}{t^2}=lim_{tto {0}} frac {2 sin ^2(t/2)}{4(t/2)^2}={1over 2}$$
answered yesterday
user376343
2,2331716
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Set $t=frac pi2 - x,$
$$lim_{xto {piover 2}} frac {1-sin x}{(fracpi2 -x)^2}=lim_{tto {0}} frac {1-cos t}{t^2}=lim_{tto {0}} frac {2 sin ^2(t/2)}{4(t/2)^2}={1over 2}$$
answered yesterday
user376343
2,2331716
Set $t=frac pi2 - x,$
$$lim_{xto {piover 2}} frac {1-sin x}{(fracpi2 -x)^2}=lim_{tto {0}} frac {1-cos t}{t^2}=lim_{tto {0}} frac {2 sin ^2(t/2)}{4(t/2)^2}={1over 2}$$
answered yesterday
user376343
2,2331716
edited yesterday
answered yesterday
user376343
2,2331716
answered yesterday
user376343
2,2331716
answered yesterday
user376343
2,2331716
2,2331716
add a comment |
add a comment |
up vote
0
down vote
We have that
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }
=lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 } frac {1+sin x}{1+sin x}
=lim_{xto frac{pi}2} frac {1}{1+sin x}frac {sin^2left(fracpi2 -xright)}{left(fracpi2 -xright)^2 } =frac12$$
answered yesterday
gimusi
86k74292
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
Got it, thank you!
– Lukáš Frajt
yesterday
add a comment |
up vote
0
down vote
We have that
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }
=lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 } frac {1+sin x}{1+sin x}
=lim_{xto frac{pi}2} frac {1}{1+sin x}frac {sin^2left(fracpi2 -xright)}{left(fracpi2 -xright)^2 } =frac12$$
answered yesterday
gimusi
86k74292
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
Got it, thank you!
– Lukáš Frajt
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
We have that
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }
=lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 } frac {1+sin x}{1+sin x}
=lim_{xto frac{pi}2} frac {1}{1+sin x}frac {sin^2left(fracpi2 -xright)}{left(fracpi2 -xright)^2 } =frac12$$
answered yesterday
gimusi
86k74292
We have that
$$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }
=lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 } frac {1+sin x}{1+sin x}
=lim_{xto frac{pi}2} frac {1}{1+sin x}frac {sin^2left(fracpi2 -xright)}{left(fracpi2 -xright)^2 } =frac12$$
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
answered yesterday
gimusi
86k74292
86k74292
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
Got it, thank you!
– Lukáš Frajt
yesterday
add a comment |
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
Got it, thank you!
– Lukáš Frajt
yesterday
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
@LukášFrajt The simpler way is note that $$lim_{xto frac{pi}2} frac {1-sin x}{left(fracpi2 -xright)^2 }=lim_{xto frac{pi}2} frac {1-cos left(fracpi2 -xright)}{left(fracpi2 -xright)^2 }=frac12$$ by standard limits.
– gimusi
yesterday
1
1
Got it, thank you!
– Lukáš Frajt
yesterday
Got it, thank you!
– Lukáš Frajt
yesterday
add a comment |
Lukáš Frajt is a new contributor. Be nice, and check out our Code of Conduct.
Lukáš Frajt is a new contributor. Be nice, and check out our Code of Conduct.
Lukáš Frajt is a new contributor. Be nice, and check out our Code of Conduct.
Lukáš Frajt is a new contributor. Be nice, and check out our Code of Conduct.
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Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
yesterday
Why isn't this trivially $0$? You have a constant in the denominator (which can be removed) and the numerator is clearly $0$.
– David G. Stork
yesterday
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
yesterday
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
yesterday