limit of integration
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We know $1leq m<n-1$, now I want to see what happens to following limit:
$$lim_deltarightarrow 0^+int_delta^2^delta t^-m-2sin^n-1(t) dt$$
I guess we don't have explicit result for the integration. So I think $sin(t)sim t$ as $trightarrow 0^+$, thus we see above is equivalent to
$$lim_deltarightarrow 0^+int_delta^2^deltat^n-m-3 dt=lim_deltarightarrow 0^+frac1n-m-2(delta^n-m-2-delta^2(n-m-2))=0$$
as long as $n-m-2> 0.$ Is my analysis correct?
calculus real-analysis
asked Aug 9 at 17:42
H-H
30118
add a comment |Â
up vote
0
down vote
favorite
We know $1leq m<n-1$, now I want to see what happens to following limit:
$$lim_deltarightarrow 0^+int_delta^2^delta t^-m-2sin^n-1(t) dt$$
I guess we don't have explicit result for the integration. So I think $sin(t)sim t$ as $trightarrow 0^+$, thus we see above is equivalent to
$$lim_deltarightarrow 0^+int_delta^2^deltat^n-m-3 dt=lim_deltarightarrow 0^+frac1n-m-2(delta^n-m-2-delta^2(n-m-2))=0$$
as long as $n-m-2> 0.$ Is my analysis correct?
calculus real-analysis
asked Aug 9 at 17:42
H-H
30118
Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We know $1leq m<n-1$, now I want to see what happens to following limit:
$$lim_deltarightarrow 0^+int_delta^2^delta t^-m-2sin^n-1(t) dt$$
I guess we don't have explicit result for the integration. So I think $sin(t)sim t$ as $trightarrow 0^+$, thus we see above is equivalent to
$$lim_deltarightarrow 0^+int_delta^2^deltat^n-m-3 dt=lim_deltarightarrow 0^+frac1n-m-2(delta^n-m-2-delta^2(n-m-2))=0$$
as long as $n-m-2> 0.$ Is my analysis correct?
calculus real-analysis
asked Aug 9 at 17:42
H-H
30118
We know $1leq m<n-1$, now I want to see what happens to following limit:
$$lim_deltarightarrow 0^+int_delta^2^delta t^-m-2sin^n-1(t) dt$$
I guess we don't have explicit result for the integration. So I think $sin(t)sim t$ as $trightarrow 0^+$, thus we see above is equivalent to
$$lim_deltarightarrow 0^+int_delta^2^deltat^n-m-3 dt=lim_deltarightarrow 0^+frac1n-m-2(delta^n-m-2-delta^2(n-m-2))=0$$
as long as $n-m-2> 0.$ Is my analysis correct?
calculus real-analysis
asked Aug 9 at 17:42
H-H
30118
edited Aug 9 at 20:23
asked Aug 9 at 17:42
H-H
30118
asked Aug 9 at 17:42
H-H
30118
asked Aug 9 at 17:42
H-H
30118
30118
Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24
add a comment |Â
Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24
Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24
add a comment |Â
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Yes, it is correct.
– uniquesolution
Aug 9 at 19:40
@uniquesolution Thank you, I also convinced myself by expanding $sin(t)$, higher order terms would vanish.
– H-H
Aug 9 at 20:24