# U: Prove Limit $\lim_{x{\rightarrow}c}\sqrt{x}=\sqrt{c}$

• kreil
You have 2 things to control, the top and the bottom. So you have some freedom in finding things you can plug in to make sure that both top and bottom are controlled by |lambda| < delta.In summary, to show that \lim_{x{\rightarrow}1}\frac{x^2-x+1}{x+1}=\frac{1}{2}, you need to find a way to bound the expression |\frac{x^2-x+1}{x+1}-\frac{1}{2}| by controlling both the top and bottom terms. Using the hint of letting x=1+lambda, where lambda approaches 0 as x approaches 1, you can show that the expression can be rewritten as
kreil
Gold Member
Show $$\lim_{x{\rightarrow}c}\sqrt{x}=\sqrt{c}$$

In other words, $$\forall \epsilon > 0 \exists \delta > 0 s.t. |x-c|< \delta \implies |\sqrt{x}-\sqrt{c}|< \epsilon$$

So $$0<|x-c|<\delta \implies |\sqrt{x}-\sqrt{c}|=|\frac{x-c}{\sqrt{x}+\sqrt{c}}|$$

and this is where I am stuck, a hint the teacher gave was "bound the denominator" but I am not sure how to do that...can anyone help?

Josh

You want to show:

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c} } < \epsilon$$

The way these things work is that you want to find something bigger than the LHS that is still smaller than RHS, because then the LHS is definitely smaller than the RHS. So keep replacing the LHS by something bigger and bigger until it's in a form that is easy to compare to the RHS. What can you replace the denominator by (that's independent of x) to get something larger than the LHS?

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{1}<\delta=\epsilon$$

Can I do that and just choose delta = epsilon?

No, that isn't true. If x=c=1/16, then the denominator is 1/2, and so the LHS is not less than the middle. The bound will depend on c.

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{3\sqrt{c}}$$

then this should work since if x=c=1/16, the LHS is 1/2 and the RHS is 3/4

can I then continue this so that

$$\frac{|x-c|}{3\sqrt{c}}<3\sqrt{c}\delta=\epsilon$$ ...??

No. First of all, you can't show something works with a single example. But it doesn't even work for that one. Remember, if you want to increase a fraction, you have to decrease its denominator. So you have to find something that is less than $\sqrt{x}+\sqrt{c}$.

so how about just squareroot of c?

Yea, that's less than $\sqrt{x}+\sqrt{c}$. So what's your answer?

In other words, choose $$\delta=\frac{\epsilon}{\sqrt{c}}$$

$$\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{\sqrt{c}}<\sqrt{c} \delta = \epsilon$$

Last edited:
Looks good.

Another limit question

I need to show that $$\lim_{x{\rightarrow}a}x^3=a^3$$

then $$0<|x-a|<\delta \implies |x^3-a^3|< \epsilon$$

I'm having a lot of trouble getting the cubed terms to be less than an expression with |x-a|. Should I try turning the cubed terms into a fraction? I need a direction to go in.

Josh

x^3 - a^3 = (x - a)(x^2 + ax + a^2).

Also you could use |x|, |a| < |a| + e. That turns the second factor into something less than 3(a + e)^2

Yet Another E-d proof

I got that one now, thanks.

Now this one is giving me a problem...

Show $$\lim_{x{\rightarrow}1}\frac{x^2-x+1}{x+1}=\frac{1}{2}$$

i.e. show $$\forall \epsilon > 0 \exists \delta > 0 s.t. |x-1|< \delta \implies |\frac{x^2-x+1}{x+1}-\frac{1}{2}|<\epsilon$$

The teacher says to let $$x=1+ \lambda$$ so that as $$x \rightarrow 1, \lambda \rightarrow 0$$. I wasn't quite sure how to properly incorporate this into the proof, because it gave me problems choosing delta:

$$|\frac{x^2-x+1}{x+1}-\frac{1}{2}|=|\frac{2x^2-3x+1}{2(x+1)} |=|\frac{(1+ \lambda -1)(2+2 \lambda -1)}{2(1+ \lambda +1)}|=|\frac{(\lambda)(2 \lambda +1)}{2( \lambda + 2)}=0< \epsilon$$

I know that this can't be right since it doesn't even require me to choose a delta, so it doesn't really satisfy the definition. Any help incorporating the hint will be helpful.

Last edited:
lambda is not equal to 0, lambda is some constant. You want to show that when 0 < |lambda| < delta for some delta, your expression on the right is less than epsilon.

## 1. What is a limit?

A limit is the value that a function approaches as its input approaches a certain value. In other words, it is the value that the function gets closer and closer to, but may never actually reach, as the input gets closer and closer to a specific value.

## 2. How do you prove a limit?

In order to prove a limit, you must show that for any small positive number, there exists a corresponding small interval around the input value such that the output values of the function fall within that interval. Essentially, you must demonstrate that the function gets arbitrarily close to the limit value as the input value gets closer to the specified value.

## 3. What is the limit of the square root function as the input approaches a specific value?

The limit of the square root function as the input approaches a specific value is the square root of that value. In other words, as the input gets closer and closer to the specified value, the output values of the function also get closer and closer to the square root of that value.

## 4. How is the limit of the square root function different from the limit of other functions?

The limit of the square root function is different from the limit of other functions in that it is a one-sided limit, meaning that it only considers the behavior of the function as the input approaches the specified value from one direction (either from the left or the right). Other functions may have two-sided limits, where the behavior of the function is considered from both directions.

## 5. Why is proving the limit of the square root function important?

Proving the limit of the square root function is important because it is a fundamental concept in calculus and is used to solve many real-world problems. It also helps to understand the behavior of functions and their limits, which can lead to further mathematical discoveries and applications.

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