Posted by Bruce on .
What is the lim of the square root of x times sin of (1/x) as x approaches infinity

Calc 2 
Damon,
x^.5 * sin (1/x)
as 1/x > 0 , sin (1/x) = 1/x
x^.5 * x^1 = x^.5 = 1/sqrt x
which = 0 as x > infinity 
Calc 2 
Steve,
or, by this time you have probably seen L'Hospital's Rule, so
limit √x sin(1/x)
if u = 1/x, then we have
limit sin(u)/√u
= limit cos(u)/(1/2√u)
= limit 2√u cos(u) as u>0
= 0