How to solve that Limit without graphing ? #limx->0^+# #1/x# = #oo# #limx->0^-# #1/x# #-oo#

1 Answer
Apr 20, 2018

Plug in a number for #x# such as #0.0000001# which is #0^+#

Explanation:

Given: #lim x-> 0^+ 1/x = oo; " "lim x-> 0^- 1/x = -oo#

Without graphing you can plug in a value that approaches #0^+# and #0^-#.

#x->0^+# means approach from the right side (positive) of zero:
Let #x = .01; " " 1/x = 100#

Let #x = .0001; " " 1/x = 10,000#

Let #x = .000001; " " 1/x = 1,000,000#

So, as #x ->0^+ 1/x = oo#

#x->0^-# means approach from the left side (negative) of zero:
Let #x = -.01; " " 1/x = -100#

Let #x = -.0001; " " 1/x = -10,000#

Let #x = -.000001; " " 1/x = -1,000,000#

So, as #x->0^-# #1/x = -oo#