- #1

cbarker1

Gold Member

MHB

- 349

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Dear Everybody,

I am having trouble to determine the value of delta when c is strictly greater than 0. Here is the work:

The Problem: Find the Limit or prove that the limit DNE.

$\lim_{{x}\to{c}}\sqrt{x} for c\ge0$

Proof:

Case I: if c>0.

Let $\varepsilon>0$ Then there exists $\delta>0$ such that if $x\ne c$ and $\left| x-c \right|>\delta$, then what needs to show is that $\left| \sqrt{x}-\sqrt{c} \right|>\varepsilon$ is true.

$\left| \sqrt{x}-\sqrt{c} \right|$

=$\left| \sqrt{x}-\sqrt{c} \right|\left| \sqrt{x}+\sqrt{c} \right|\frac{1}{\left| \sqrt{x}+\sqrt{c} \right|}$

$\frac{\left| x-c \right|}{\left| \sqrt{x}+\sqrt{c} \right|}$

Here is where I am lost?

Thanks,

CBarker1

I am having trouble to determine the value of delta when c is strictly greater than 0. Here is the work:

The Problem: Find the Limit or prove that the limit DNE.

$\lim_{{x}\to{c}}\sqrt{x} for c\ge0$

Proof:

Case I: if c>0.

Let $\varepsilon>0$ Then there exists $\delta>0$ such that if $x\ne c$ and $\left| x-c \right|>\delta$, then what needs to show is that $\left| \sqrt{x}-\sqrt{c} \right|>\varepsilon$ is true.

$\left| \sqrt{x}-\sqrt{c} \right|$

=$\left| \sqrt{x}-\sqrt{c} \right|\left| \sqrt{x}+\sqrt{c} \right|\frac{1}{\left| \sqrt{x}+\sqrt{c} \right|}$

$\frac{\left| x-c \right|}{\left| \sqrt{x}+\sqrt{c} \right|}$

Here is where I am lost?

Thanks,

CBarker1

Last edited: