proving that $\lim \frac{x^2 - 1}{x - 1} =2$ as $x \rightarrow 1.$ by definition

Question

proving that $\lim \frac{x^2 - 1}{x - 1} =2$ as $x \rightarrow 1.$ by limit definition.

My answer: take $\delta \leq \epsilon$, am I correct?

Answer 1

am I correct?

Your idea is correct however you'll have to write much more.

You first have to repeat the definition used by you:

The definition of a $\lim\limits_{x \to a}$ ... is:

For every value $\epsilon > 0$ there is a value $\delta > 0$ that ...

Otherwise the other people will ask you:

What is $\delta$ and what is $\epsilon$?

At least one user already did this in the comment.

Then you claim that any $\delta \leq \epsilon$ satisfies this condition so for every $\epsilon$ such a $\delta$ exists.

You also have to prove this.

Otherwise I will claim that there is some value $x \in [0.6,1.4]$ which satisfies $\frac{x^2-1}{x-1}=100000$. If you don't prove your claim I won't prove mine either.

If my claim was right this would mean that the value $\delta = 0.5$ would not fulfill the condition for $\epsilon = 1$ so your claim that any $\delta\leq\epsilon$ fulfils the conditions would be wrong.