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

Question

Isabella Ackerman · Accepted Answer

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

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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Christopher Allers · Answer

undefined To prove that the function g(x) = x^3 / x is continuous everywhere except at x = 0, we need to show that it satisfies the conditions for continuity.

First, let's consider the function g(x) when x ≠ 0. In this case, we can simplify the function by canceling out the x in the numerator and denominator, resulting in g(x) = x^2.

The function x^2 is a polynomial function, and all polynomial functions are continuous for all real numbers. Therefore, g(x) = x^2 is continuous for all x ≠ 0.

Now, let's examine the behavior of g(x) as x approaches 0. We can rewrite the function as g(x) = x^3 / x = x^2 when x ≠ 0.

As x approaches 0, the function g(x) approaches 0 as well. This can be seen by evaluating the limit of g(x) as x approaches 0, which is lim(x→0) x^2 = 0.

Since the limit of g(x) as x approaches 0 exists and is equal to the value of g(0), which is also 0, we can conclude that g(x) is continuous at x = 0.

In summary, the function g(x) = x^3 / x is continuous everywhere except at x = 0, as it satisfies the conditions for continuity for all x ≠ 0 and is also continuous at x = 0.