Limxtends to infinityx/modulus of x..?

1 Answer
Jim H
Mar 28, 2018

Please see below.

Explanation:

I am interpreting "#x# tends to infinity" as meaning "#x# increases without bound".

#x/absx = {(1,"if",x > 0),(-1,"if",x < 0):}#

Therefore #lim_(xrarroo) x/absx = 1#

Bonus Answer

As #x# tends to minus infinity (as #x# decreases without bound), we have

#lim_(xrarr-oo) x/absx = -1#

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