Find Limit of $\sqrt{x}$ as $x\to c$, $c\ge 0$

[cbarker1]

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

 

[Evgeny.Makarov]

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.

That's not the definition of limit. And it's not the negation of the definition, either.

 

[math771]

Dear CBarker1,

I can help you with your problem. The key to solving this limit is to use the definition of a limit and the properties of square roots.

First, let's rewrite the expression $\left| \sqrt{x}-\sqrt{c} \right|$ as $\frac{\left| x-c \right|}{\sqrt{x}+\sqrt{c}}$. This is because we know that $\sqrt{x}+\sqrt{c}$ is a factor of $\left| \sqrt{x}-\sqrt{c} \right|$.

Next, we can use the triangle inequality to rewrite $\left| x-c \right|$ as $\left| \sqrt{x}-\sqrt{c} \right| + \left| \sqrt{c} \right|$. This is because $\left| x-c \right| = \left| \sqrt{x}-\sqrt{c} \right| + \left| \sqrt{c} \right|$.

Now, we can substitute this into our original expression and get $\frac{\left| \sqrt{x}-\sqrt{c} \right| + \left| \sqrt{c} \right|}{\sqrt{x}+\sqrt{c}}$.

Finally, we can use the fact that $\left| \sqrt{x}-\sqrt{c} \right| < \varepsilon$ and $\left| \sqrt{c} \right| < \varepsilon$ to show that $\frac{\left| \sqrt{x}-\sqrt{c} \right| + \left| \sqrt{c} \right|}{\sqrt{x}+\sqrt{c}} < \frac{\varepsilon + \varepsilon}{\sqrt{x}+\sqrt{c}} = \frac{2\varepsilon}{\sqrt{x}+\sqrt{c}}$.

Now, we can choose $\delta = \frac{2\varepsilon}{\sqrt{x}+\sqrt{c}}$ and this will satisfy the definition of the limit. Therefore, the limit exists and is equal to $\sqrt{c}$.

I hope this helps! Let me know if you have any further questions.