# Simple limit

1. Nov 27, 2007

### Joza

Could someone please, step by step, compute the limit as x goes to 0, x >0, of:

1/(2(x^1/2))

I am confused about it. Thanks.

2. Nov 27, 2007

### rock.freak667

3. Nov 27, 2007

If you know 1/x goes to $$\infty$$ then you can use a comparison test $$\frac{1}{\sqrt{x}} \geq \frac{1}{x}$$ for 0 < x < 1.

Otherwise you have to apply the definition of a limit tending to infinity..

For $$0 < x < \frac{1}{M^2}$$ you have $$\frac{1}{\sqrt{x}} > M$$.

4. Nov 28, 2007

### HallsofIvy

It's not clear what you mean by "compute" the limit. Since x is in the denominator,I would think it obvious that, as x goes to 0, the fraction goes to infinity.

5. Nov 28, 2007

### Office_Shredder

Staff Emeritus
Then $$\sqrt{x} \leq x$$ for 0<x<1 which means $$x \leq x^2$$ which gives $$1 \leq x$$

6. Nov 28, 2007

Correction noted: 1/sqrt(x) <= 1/x. Comparison test doesn't work.

7. Nov 28, 2007

### bomba923

$$\lim \limits_{x \to 0^ + } \frac{1}{{2\sqrt x }} = 0\because \forall \varepsilon > 0,\;\exists \delta > 0:0 < x < \delta \to \frac{1} {{2\sqrt x }} < \varepsilon ,\;{\text{simply choose }}\delta = \frac{1} {{4\varepsilon ^2 }}$$