How do you find #lim 1/x^2-1/x# as #x->0#?

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Answer 1 · 2017-01-08T13:46:28

#lim_(x->0) 1/(x^2)-1/x = +oo#

You can simplify the function in this way:

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

Now:

#lim_(x->0) 1/(x^2) = +oo#

#lim_(x->0) (1-x) = 1#

So:

#lim_(x->0) 1/(x^2)-1/x = +oo#

graph{1/x^2-1/x [-10, 10, -5, 5]}