# Improper integral limit

1. Jul 22, 2012

### marellasunny

$$\int_0^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x$$
=

$$\lim_{\varepsilon \to 0+}\int_\varepsilon^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x$$

My question is about the usage of 0+ in the limit.(I evaluated the integrals and arrived at the part where I substitute upper and lower limits.)
Did the author deliberately choose to use $$\lim_{\varepsilon \to 0+}$$ instead of 0 or 0- so that any imaginary numbers arising from the expression $$2\sqrt{x}$$ do not arise?
Or is there any other reason?
Thanks.

2. Jul 22, 2012

### DonAntonio

I don't know what author you're talking about, but taking that limit is what has to be done simply by the definition of improper

integral with one of the limits being a point of unboundness for the function...

DonAntonio

3. Jul 22, 2012

### marellasunny

Yes,i understand this case.But,what if I had a case of a function best described by limit->0-?
Will I not have a problem when I substitute 0- into the square root?

I can't exactly describe the function,I mean it for some arbitrary function have variable 'x' under the square root and me having to apply limit->0-.Wont this give rise to a imaginary number?

4. Jul 22, 2012

### Number Nine

Taking the limit from below would result in the expression being undefined.

5. Jul 22, 2012

### DonAntonio

I'm not completely sure I follow you, but the function in the integral is defined only for positive real numbers: whatever you want

to do with it will have to comply with this restriction. Thus, there is not meaning to the expression
$$\lim_{x\to 0^-}\sqrt x$$
as it assumes the existence of the square roots of negative numbers within the real numbers, which is absurd.

DonAntonio