Continuous Function over its natural domain (for f(x)=1/x, but not for g(x)=1/x^2 ??)

[shanghong]

Thomas Calculus 14th edition, p93

It says that f(x)=1/x is a continuous function over its natural domain because x=0 is not in the domain of the function.

However, it also says that g(x)=1/(x^2) has an infinite discontinuity when x→0, the function → ∞.

My question:
If f(x)=1/x is a continuous function, then I must say that the function g(x)=1/x^2 is also a continuous function over its natural domain because x can never be zero naturally in order to be a valid function. Why does the book say that g(x)=1/(x^2) has an infinite discontinuity?

Thanks

 

[pka]

Thomas Calculus 14th edition, p93
It says that f(x)=1/x is a continuous function over its natural domain because x=0 is not in the domain of the function.
However, it also says that g(x)=1/(x^2) has an infinite discontinuity when x→0, the function → ∞.
My question: If f(x)=1/x is a continuous function, then I must say that the function g(x)=1/x^2 is also a continuous function over its natural domain because x can never be zero naturally in order to be a valid function. Why does the book say that g(x)=1/(x^2) has an infinite discontinuity?

I have the 9th edition so this may not help you. On page 89 they define what it means to say a function is continuous on a set. On its natural domain the function \(\displaystyle f(x)=\dfrac{1}{x}\) is clearly continuous at each point. I cannot find a discussion for \(\displaystyle g(x)=\dfrac{1}{x^2}\)

Perhaps you might consider this.
\(\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x^2}}}} \right) = \infty \) whereas \(\displaystyle \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{{{x}}}} \right) \ne \infty \) because \(\displaystyle \mathop {\lim }\limits_{x \to 0^+} \left( {\frac{1}{{{x}}}} \right) = \infty \) and \(\displaystyle \mathop {\lim }\limits_{x \to 0^{\bf-}} \left( {\frac{1}{{{x}}}} \right) = -\infty \)

 

[lev888]

Could you post the author's definitions of
1. continuous function
2. infinite discontinuity?

I'd say they are both continuous over their domains and have an infinite discontinuity at 0.

 

[Dr.Peterson]

Thomas Calculus 14th edition, p93

It says that f(x)=1/x is a continuous function over its natural domain because x=0 is not in the domain of the function.

However, it also says that g(x)=1/(x^2) has an infinite discontinuity when x→0, the function → ∞.

My question:
If f(x)=1/x is a continuous function, then I must say that the function g(x)=1/x^2 is also a continuous function over its natural domain because x can never be zero naturally in order to be a valid function. Why does the book say that g(x)=1/(x^2) has an infinite discontinuity?

Judging by the bits of Thomas that I was able to find online, I think you are imagining a contradiction that is not there. He says that 1/x is continuous over its domain, and that 1/x^2 has an infinite discontinuity; he does not say that 1/x doesn't have an infinite discontinuity, or that 1/x^2 is not continuous over its domain. Both are true of both. They are just different ways of looking at the same phenomenon, also called a vertical asymptote.

 

[stapel]

Judging by the bits of Thomas that I was able to find online, I think you are imagining a contradiction that is not there. He says that 1/x is continuous over its domain, and that 1/x^2 has an infinite discontinuity; he does not say that 1/x doesn't have an infinite discontinuity, or that 1/x^2 is not continuous over its domain. Both are true of both....

But, thinking of myself as a student encountering these terms and concepts for the first time, I think I'd find this presentation very confusing! :shock:

 

[Dr.Peterson]

But, thinking of myself as a student encountering these terms and concepts for the first time, I think I'd find this presentation very confusing! :shock:

I fully agree, assuming it is just as described. I considered saying that, but chose not to without having read the whole thing.