Limits proof

x

I have only had a very brief look, but what you've written looks mostly okay. The following line is a bit confusing as you've clearly done some working that isn't shown, but I presume you've written it out in my detail on paper!:

"Taking |root(6-x) - 2| we times both sides by root(6-x) + 2 giving| (2-x)/(root6-x)+2| = |x-2| / | (root 6-x) + 2|"

Also, there's a typo in this line:

"So |x-2| / | (root 6-x) + 2| < |x-2| so let delta = epsilon." It should obviously have $<\epsilon$ rather than $<|x-2|$.

Yeh, I meant epsilon.

Ther trouble I had was preventing root(6-x) from being negative.

Was it ok to say in the definition xER \ {x&lt;/6} ???

Yeah, maybe just say at the start of your answer that $\sqrt{6-x}$ is not defined on $\mathbb{R}$ when x>6, hence you will apply the definition for x<=6.

I'm struggling a bit with how to do this..

A typical trick, which I'm sure you'll have seen used in different examples in your book/lectures, is to say just change the notation a little and write something like:

"Now we can suppose (For all $\epsilon _1 >0$) (There exists delta >0) (For all x E X)
0<|x-a|<delta implies $|f(x) - L| < \epsilon _1$."

Maybe go back and see how those kinds of proofs were done, and think about how you can apply similar ideas to this question?

I've still not managed this

You've just changed epsilon to epsilon 1?

Can you give me more of a hint!

I've still not managed this

You've just changed epsilon to epsilon 1?

Can you give me more of a hint!

The next stage is to consider $|2f(x) - 2L|$. You know that there exists a delta such that $|f(x) - L| < \epsilon _1$, so you will be able to find a bound for $|2f(x) - 2L|$ as well (using the same delta).

I did this:

Suppose (For all epsilon &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |f(x) - L| &lt; epsilon

As epsilon &gt; 0, epsilon/2 &gt; 0 so

(For all epsilon/2 &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |f(x) - L| &lt; epsilon/2


|f(x)-L| &lt; epsilon / 2 can be written |2f(x) - 2L| &lt; epsilon

So (For all epsilon &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |2f(x) - 2L| &lt; epsilon



I kind of know this isn't sufficient. Could you tweak it for me!