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

• MHB
• cbarker1
In summary: Expert SummarizerIn summary, the expert summarizer explains how to solve the limit problem of $\lim_{{x}\to{c}}\sqrt{x}$ for $c\ge0$ by using the definition of a limit and the properties of square roots. The expert also provides the necessary steps to show that the limit exists and is equal to $\sqrt{c}$.
cbarker1
Gold Member
MHB
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

Last edited:
Cbarker1 said:
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.

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.

## 1. What is the limit of $\sqrt{x}$ as $x$ approaches $c$ from the right?

The limit of $\sqrt{x}$ as $x$ approaches $c$ from the right is equal to the square root of $c$. This can be written as $\lim_{x\to c^+}\sqrt{x}=\sqrt{c}$.

## 2. How do I find the limit of $\sqrt{x}$ as $x$ approaches $c$?

To find the limit of $\sqrt{x}$ as $x$ approaches $c$, you can simply plug in the value of $c$ into the function. If the function is continuous at $c$, the limit will be equal to the value of the function at $c$. If the function is not continuous at $c$, you can use algebraic techniques or a graph to determine the limit.

## 3. Can the limit of $\sqrt{x}$ as $x$ approaches $c$ be negative?

No, the limit of $\sqrt{x}$ as $x$ approaches $c$ cannot be negative. This is because the square root function is only defined for non-negative values, so as $x$ gets closer to $c$, the output of the function will also approach non-negative values.

## 4. What is the limit of $\sqrt{x}$ as $x$ approaches $c$ from the left?

The limit of $\sqrt{x}$ as $x$ approaches $c$ from the left is also equal to the square root of $c$. This can be written as $\lim_{x\to c^-}\sqrt{x}=\sqrt{c}$. This is because the square root function is continuous from the left.

## 5. Can the limit of $\sqrt{x}$ as $x$ approaches $c$ be undefined?

Yes, the limit of $\sqrt{x}$ as $x$ approaches $c$ can be undefined. This can occur if the function has a vertical asymptote at $c$ or if the function is not defined at $c$. In these cases, the limit does not exist.

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