What is #lim_ (xrarr0) (1/x -1/x^2-x)#?

1 Answer
Nov 5, 2015

Since both limits from left and right are #-infty#, the limit is also #-infty#.

Explanation:

Since #x->0#, the last term #-x# is irrelevant. So, let's focus on the rest.

We have that, if #x->0^-#, then #1/x -> -infty#, and so does #-1/x^2#. So we have #-infty-infty = -infty#.

On the other hand, if #x->0^+#, then we are in a #infty-infty# form, which we can't solve directly.

So, we must sum the two fractions:

#1/x-1/x^2 = (x^2-x)/x^3 = (x-1)/x^2#

And now the limit as #x->0^+# of this quantity is #(-1)/0^+ = -infty#,