What is the limit of 1/(2(x^1/2)) as x approaches 0 from the positive side?

[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.

 

[rock.freak667]

Read here

 

[rudinreader]

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 $$\frac{1}{\sqrt{x}} > M$$.

 

[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.

 

[Office_Shredder]

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.

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

 

[rudinreader]

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

 

[bomba923]

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

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