# Does lim_(ararr0^+)(1/a^2)+lim_(brarr0-)(-1/b^2) diverge?

## I was solving ${\int}_{-} {1}^{1} \left(\frac{1}{x} ^ 3\right) \mathrm{dx}$ and my teacher said it diverges. I was wondering if you can do some kind of cancelling to get ${\lim}_{a \rightarrow {0}^{+}} \left(\frac{1}{a} ^ 2\right) + {\lim}_{b \rightarrow 0 -} \left(- \frac{1}{b} ^ 2\right)$ equal to zero.

##### 1 Answer
Jan 14, 2018

Please see below.

#### Explanation:

The standard (basic) definition of convergence of an improper integral tells us that

If $f$ is a function that is not defined at $0$, then

${\int}_{-} {a}^{a} f \left(x\right) \mathrm{dx}$ converges if and only if both ${\int}_{-} {a}^{0} f \left(x\right) \mathrm{dx}$ and ${\int}_{0}^{a} f \left(x\right) \mathrm{dx}$ converge.

There is another quantity of interest, called the Cauchy Principle Value. For the integral you are working on,
the Cauchy Principle Value is

${\lim}_{\epsilon \rightarrow {0}^{+}} \left[{\int}_{-} {1}^{0 - \epsilon} \frac{1}{x} ^ 3 \mathrm{dx} + {\int}_{0 + \epsilon}^{1} \frac{1}{x} ^ 3 \mathrm{dx}\right] = 0$

The standard (or Basic) definition requires us to take the two limit separately. The Cauchy principal value takes the limit of the sum.

As for you specific question, I am not confident about any definition of convergence. that would allow us to answer the question, "Does ${\lim}_{a \rightarrow {0}^{+}} \left(\frac{1}{a} ^ 2\right) + {\lim}_{b \rightarrow 0 -} \left(- \frac{1}{b} ^ 2\right)$ converge?"

I can say that in the extended real numbers this evaluates to $\infty + \left(- \infty\right)$ which is not defined in the extended reals. So I would call that divergent.