Evaluate the limit? # lim_(x rarr 0) x^2sin(1/x) #

Question

1034 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-14T17:08:44

# lim_(x rarr 0) x^2sin(1/x) = 0#

We want to find:

# L = lim_(x rarr 0) x^2sin(1/x) #

graph{x^2sin(1/x) [-0.3268, 0.3302, -0.1632, 0.1654]}

Graphically, it looks as though #L=0# so let us see if we can prove this analytical:

Let #z = 1/x# then as # x rarr 0 => z rarr oo#

So then, the limit can be written:

# L = lim_(z rarr oo) (1/z)^2sinz #
# \ \ = lim_(z rarr oo) (sinz)/z^2 #
# \ \ = 0 #

As #|sin(z)| le 1# and #1/z^2 rarr 0# as #z rarr oo#