How do you find the limit of [ (√x) - (√sin(x)) ] / x^(5/2) as x goes to 0

lim(x-->0) [ √x - √sin(x) ] / x^(5/2)=
=lim(x-->0) [ √x - √sin(x) ] / [x²√x]=
=lim(x-->0) [ 1 - √(sin(x)/x) ] / x²=
=lim(x-->0) [ 1 - √(sin(x)/x) ][ 1 + √(sin(x)/x) ] / (x²[ 1 + √(sin(x)/x) ])=
=lim(x-->0) [ 1 - (sin(x)/x) ]/ x² * lim(x-->0) 1/[ 1 + √(sin(x)/x) ]=
=lim(x-->0) [ 1 - (sin(x)/x) ]/ x² * 1/[ 1 + √1 ]=
=½lim(x-->0) [ 1 - (sin(x)/x) ]/ x² =
=½lim(x-->0) [ x - sin(x) ]/ x³ =
(L'Hospital's Rule)
=½lim(x-->0) ( x - sin(x) )`/ (x³)` =
=½lim(x-->0) ( 1 - cos(x) )/ (3x²) =
(L'Hospital's Rule)
=½lim(x-->0) ( 1 - cos(x) )`/ (3x²)` =
=½lim(x-->0) sin(x)/ (6x) =
=1/12
http://www.wolframalpha.com/input/?i=lim…

by logical thinking:

√sin(x) is a period which never gets higher than 1

√x rises if x rises ans the other way round

because of the squareroot (√x) - (√sin(x)) must be positive

x^(5/2) rises if x decreases

the rising of x^(5/2) is higher than the decreasing of (√x) - (√sin(x)) if x comes near to 0

so the limit must be infinite