What to use for $lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$
$begingroup$
Determine $$lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$$
I know the three convergence theorems, but to no avail:
$1.$ Monotone Convergence:
The series $f_{n}(x):=frac{nsin{x}}{1+n^2sqrt{x}}$ is not monotonic increasing on $[0,1]$, so the conditions are not met.
$2.$ Fatou:
Note: $liminf_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dxgeqint_{[0,1]}liminf_{nto infty}frac{nsin{x}}{1+n^2sqrt{x}}dx=0$
Which aids us no further.
$3.$ Dominated Convergence Theorem:
The only function $h$ to fulfill $|f_{n}|leq h, forall n in mathbb N$ that comes to mind is $|frac{nsin{x}}{1+n^2sqrt{x}}|=frac{nsin{x}}{n^2sqrt{x}}=frac{sin{x}}{nsqrt{x}}leq frac{sin{x}}{sqrt{x}}=:h(x).$
But how do I show $h$ is $in mathcal{L}^1$?
Is there anything I am missing? Any guidance is greatly appreciated.
real-analysis measure-theory convergence
edited Dec 29 '18 at 20:03
Namaste
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
$endgroup$
add a comment |
$begingroup$
Determine $$lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$$
I know the three convergence theorems, but to no avail:
$1.$ Monotone Convergence:
The series $f_{n}(x):=frac{nsin{x}}{1+n^2sqrt{x}}$ is not monotonic increasing on $[0,1]$, so the conditions are not met.
$2.$ Fatou:
Note: $liminf_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dxgeqint_{[0,1]}liminf_{nto infty}frac{nsin{x}}{1+n^2sqrt{x}}dx=0$
Which aids us no further.
$3.$ Dominated Convergence Theorem:
The only function $h$ to fulfill $|f_{n}|leq h, forall n in mathbb N$ that comes to mind is $|frac{nsin{x}}{1+n^2sqrt{x}}|=frac{nsin{x}}{n^2sqrt{x}}=frac{sin{x}}{nsqrt{x}}leq frac{sin{x}}{sqrt{x}}=:h(x).$
But how do I show $h$ is $in mathcal{L}^1$?
Is there anything I am missing? Any guidance is greatly appreciated.
real-analysis measure-theory convergence
edited Dec 29 '18 at 20:03
Namaste
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
$endgroup$
add a comment |
$begingroup$
Determine $$lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$$
I know the three convergence theorems, but to no avail:
$1.$ Monotone Convergence:
The series $f_{n}(x):=frac{nsin{x}}{1+n^2sqrt{x}}$ is not monotonic increasing on $[0,1]$, so the conditions are not met.
$2.$ Fatou:
Note: $liminf_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dxgeqint_{[0,1]}liminf_{nto infty}frac{nsin{x}}{1+n^2sqrt{x}}dx=0$
Which aids us no further.
$3.$ Dominated Convergence Theorem:
The only function $h$ to fulfill $|f_{n}|leq h, forall n in mathbb N$ that comes to mind is $|frac{nsin{x}}{1+n^2sqrt{x}}|=frac{nsin{x}}{n^2sqrt{x}}=frac{sin{x}}{nsqrt{x}}leq frac{sin{x}}{sqrt{x}}=:h(x).$
But how do I show $h$ is $in mathcal{L}^1$?
Is there anything I am missing? Any guidance is greatly appreciated.
real-analysis measure-theory convergence
edited Dec 29 '18 at 20:03
Namaste
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
$endgroup$
Determine $$lim_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dx$$
I know the three convergence theorems, but to no avail:
$1.$ Monotone Convergence:
The series $f_{n}(x):=frac{nsin{x}}{1+n^2sqrt{x}}$ is not monotonic increasing on $[0,1]$, so the conditions are not met.
$2.$ Fatou:
Note: $liminf_{nto infty}int_{[0,1]}frac{nsin{x}}{1+n^2sqrt{x}}dxgeqint_{[0,1]}liminf_{nto infty}frac{nsin{x}}{1+n^2sqrt{x}}dx=0$
Which aids us no further.
$3.$ Dominated Convergence Theorem:
The only function $h$ to fulfill $|f_{n}|leq h, forall n in mathbb N$ that comes to mind is $|frac{nsin{x}}{1+n^2sqrt{x}}|=frac{nsin{x}}{n^2sqrt{x}}=frac{sin{x}}{nsqrt{x}}leq frac{sin{x}}{sqrt{x}}=:h(x).$
But how do I show $h$ is $in mathcal{L}^1$?
Is there anything I am missing? Any guidance is greatly appreciated.
real-analysis measure-theory convergence
real-analysis measure-theory convergence
edited Dec 29 '18 at 20:03
Namaste
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
edited Dec 29 '18 at 20:03
Namaste
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
edited Dec 29 '18 at 20:03
Namaste
1
edited Dec 29 '18 at 20:03
Namaste
1
edited Dec 29 '18 at 20:03
Namaste
1
1
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
asked Dec 29 '18 at 19:56
SABOYSABOY
668311
668311
add a comment |
add a comment |
2 Answers
2
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oldest
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Use dominated convergence with $h(x)=frac{sin(x)}{x}$ for $xin(0,1]$ and $h(0)=1$.
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
You’re almost there... the map $h : x mapsto frac{sin x}{sqrt x}$ has $0$ for limit as $x to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
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$begingroup$
Use dominated convergence with $h(x)=frac{sin(x)}{x}$ for $xin(0,1]$ and $h(0)=1$.
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
Use dominated convergence with $h(x)=frac{sin(x)}{x}$ for $xin(0,1]$ and $h(0)=1$.
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
$endgroup$
add a comment |
$begingroup$
Use dominated convergence with $h(x)=frac{sin(x)}{x}$ for $xin(0,1]$ and $h(0)=1$.
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
$endgroup$
Use dominated convergence with $h(x)=frac{sin(x)}{x}$ for $xin(0,1]$ and $h(0)=1$.
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
answered Dec 29 '18 at 20:04
Ben WBen W
2,276615
2,276615
add a comment |
add a comment |
$begingroup$
You’re almost there... the map $h : x mapsto frac{sin x}{sqrt x}$ has $0$ for limit as $x to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You’re almost there... the map $h : x mapsto frac{sin x}{sqrt x}$ has $0$ for limit as $x to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
You’re almost there... the map $h : x mapsto frac{sin x}{sqrt x}$ has $0$ for limit as $x to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
You’re almost there... the map $h : x mapsto frac{sin x}{sqrt x}$ has $0$ for limit as $x to 0$. Hence can be extended by continuity on $[0,1]$ and is integrable on that interval.
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Dec 29 '18 at 20:21
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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