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 Nov 20 at 22:54
gimusi
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
add a comment |
up vote
0
down 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 Nov 20 at 22:54
gimusi
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
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
Nov 20 at 22:33
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47
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 Nov 20 at 22:54
gimusi
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
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 Nov 20 at 22:54
gimusi
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
edited Nov 20 at 22:54
gimusi
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
edited Nov 20 at 22:54
gimusi
91k74495
edited Nov 20 at 22:54
gimusi
91k74495
edited Nov 20 at 22:54
gimusi
91k74495
91k74495
asked Nov 20 at 22:31
Lukáš Frajt
81
asked Nov 20 at 22:31
Lukáš Frajt
81
asked Nov 20 at 22:31
Lukáš Frajt
81
81
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
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
Nov 20 at 22:33
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47
add a comment |
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
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
Nov 20 at 22:33
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
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
Nov 20 at 22:33
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
Nov 20 at 22:33
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47
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 Nov 20 at 22:47
user376343
2,6412819
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 Nov 20 at 22:52
gimusi
91k74495
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
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 Nov 20 at 22:47
user376343
2,6412819
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 Nov 20 at 22:47
user376343
2,6412819
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 Nov 20 at 22:47
user376343
2,6412819
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 Nov 20 at 22:47
user376343
2,6412819
edited Nov 20 at 23:00
answered Nov 20 at 22:47
user376343
2,6412819
answered Nov 20 at 22:47
user376343
2,6412819
answered Nov 20 at 22:47
user376343
2,6412819
2,6412819
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 Nov 20 at 22:52
gimusi
91k74495
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
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 Nov 20 at 22:52
gimusi
91k74495
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
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 Nov 20 at 22:52
gimusi
91k74495
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 Nov 20 at 22:52
gimusi
91k74495
answered Nov 20 at 22:52
gimusi
91k74495
answered Nov 20 at 22:52
gimusi
91k74495
answered Nov 20 at 22:52
gimusi
91k74495
91k74495
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
add a comment |
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
Is that right? we still have 0/0 in the second fraction
– Lukáš Frajt
Nov 20 at 23:07
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@LukášFrajt The second fraction is in the form $yto 0 quad frac{sin^2 y}{y^2} to 1$.
– gimusi
Nov 20 at 23:09
@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
Nov 20 at 23:10
@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
Nov 20 at 23:10
1
1
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
Got it, thank you!
– Lukáš Frajt
Nov 20 at 23:11
add a comment |
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Your equation makes no sense. $sin$ needs an argument, at the very least.
– David G. Stork
Nov 20 at 22:31
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
Nov 20 at 22:33
answer in the book is 1/2. But I cannot solve it. Also in wolfram alpha..
– Lukáš Frajt
Nov 20 at 22:34
Shouldn't be the $1$ in the denominator replaced by $x$?
– user376343
Nov 20 at 22:47