How do you prove that the function #g(x) = x^3 / x# is continuous everywhere but x=0?

1 Answer
May 7, 2016

You must prove that the limit of the function exists everywhere, accept to #x=0#


There are several ways to prove continuity, they differ in key ideas and mathematical formality.

The function f is continuous at some point c of its domain if the limit of f(x) as x approaches c through the domain of f exists and is equal to f(c). In mathematical notation, this is written as

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In detail this means three conditions: 1) first, f has to be defined at c. 2) Second, the limit on the left hand side of that equation has to exist. 3) Third, the value of this limit must equal f(c).

For our function: #f(x)=x^3/x#, any value we replace we have a value, but zero, which will give #0/0#, it is undefined. Therefore, it is not continuous.

Nonetheless, as you can see from the graph below, this function is continuous on 0 , since #f(x)=x^2#. It is just the parabola masked-out.

But the ideas herein can be used to prove that a true not continuous function is not continuous in a point, e.g. #1/x#, apply the ideas herein for 0, it goes to infinity.

graph{x^3/x [-2.5, 2.5, -1.25, 1.25]}

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