#lim_(x->-oo)(x*|(1/x)|)#?

Question

1143 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-17T01:36:08

The limit does not exist (diverges to #+oo#)

For #x<-1#, we have

#0>1/x> -1#

and hence

#|__ (1/x) __| = -1#

So, for #x < -1#, we have

#x\ |__ 1/x __| = -x#

and the limit we are seeking is

#lim_{x to -oo } x\ |__ 1/x __| = lim_{x to -oo} (-x)#

It is clear that this limit does not exist (diverges to #+oo#)

#lim_(x->-oo)(x*|__(1/x)__|)#?