What is the limit as x approaches 0 of #1/x^2#?

1 Answer
MeneerNask
Jun 10, 2018

Infinity #+oo#

Explanation:

It is easily shown, that, as #x# gets smaller, #x^2# gets smaller at an even greater rate, so #1/x^2# will be greater.

A few steps:

#x=1->x^2=1->1/x^2=1#

#x=1/2->x^2=1/4->1/x^2=4#

#x=1/100->x^2=10000->1/x^2=10000#

This means that the closer #x# goes to #0# the higher the function goes. In this case it doesn't matter whether #x->0# from the positive side or from the negative, as the square makes it al positive. By choosing smaller and smaller values of #x#, the function can reach any size you want.

Translated to "the language":

#lim_(x->0^+) 1/x^2=lim_(x->0^-) 1/x^2=lim_(x->0) 1/x^2= oo#
graph{1/x^2 [-17.75, 18.3, -1.61, 16.42]}

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