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
What is the lim of the square root of x times sin of (1/x) as x approaches infinity
2 answers
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
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
1 answer