Asked by Laura
Using L'Hospital's rule:
find, lim as x approaches (-) infinity of xsin(1/x)
find, lim as x approaches (-) infinity of xsin(1/x)
Answers
Answered by
drwls
Let u = 1/x.
The limit can then be written
lim as u -> 0 of (sin u)/u
Use L'Hopital's rule on that and you get (cos 0)/1 = 1.
I could have also taken the ratio of the derivatives of sin(1/x) and (1/x), but the other way was easier.
(-1/x^2)*cos (1/x)/(-1/x^2) = cos(1/x),
the limit of which as x -> infinity
is 1
The limit can then be written
lim as u -> 0 of (sin u)/u
Use L'Hopital's rule on that and you get (cos 0)/1 = 1.
I could have also taken the ratio of the derivatives of sin(1/x) and (1/x), but the other way was easier.
(-1/x^2)*cos (1/x)/(-1/x^2) = cos(1/x),
the limit of which as x -> infinity
is 1