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Let's say we have a f: R \to R, where lim$_x$ $_\to$ $_1$ f(x) = 2.

Then we have $f^2$: R $\to$ R where x $\to$ $(f(x))^2$. Does this function still have a limit at x $\to$ 1?

So, I understand that the limit of the product of two functions is the product of each functions limit. I also know that the limit of this function would be 4.

However, I don't know how I could prove this without using known properties of limits. I tried using the definition of a limit but, I couldn't follow it up with anything. I said that:

$\forall$ x $\in$ R, $\forall$ $\epsilon$ > 0, $\exists$ $\eta$ > 0 such that |x-1|<$\eta$ $\implies$ |$f^2$(x) - $f(a)^2$)|<$\epsilon$ , but I'm not too sure where to go from there. Am I going in the right direction? Is there something I'm missing here? Thank you for your help.

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  • $\begingroup$ I don't know if this counts as "using the properties of limits", but you could use the fact that the function $x^2$ is continuous, then $\lim x_n^2=(\lim x_n)^2$ if $x_n$ converges $\endgroup$
    – user438666
    Oct 22, 2017 at 17:13
  • $\begingroup$ Thanks for your response. Sorry but I also wanted to avoid this. I know I'm being fussy, but this particular question was on last year's exam, and I know that we haven't talked about these properties in the exam's class. I don't know if it's assumed we already know them, or if they want us to use another method. $\endgroup$
    – iaskdumbstuff
    Oct 22, 2017 at 17:15
  • $\begingroup$ It isn't clear what you are trying to avoid. Properties of limits takes in a lot of material, including it seems to me what you want to prove. I like the approach by composition of two continuous functions being continuous, but if that seems unsatisfactory, then perhaps you seek a proof that proceeds from the definition? $\endgroup$
    – hardmath
    Oct 22, 2017 at 17:24
  • $\begingroup$ I understand, my apologies. The problem is, I can only assume they want us to answer the question using methods and concepts that were taught in the course, so it's difficult for me to explain what to and what not to avoid. We did learn about composition of two continuous functions being continuous in another class, but not in that particular class, so I think they would want us to avoid that answer as well. That's why I assumed they wanted us to use the definition of a function, because that's really what we learned about when it came to limits, at least in that class. $\endgroup$
    – iaskdumbstuff
    Oct 22, 2017 at 17:34

1 Answer 1

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So the $\lim _{x\to 1}f(x)=2$.

This means that there is a value $e$ such that , provided $|x-1|\lt e, 1\lt f(x)\lt 3$

We now consider the function $\left(f(x)\right)^2-4=\left(f(x)+2\right)\left(f(x)-2\right)$

We can choose $\epsilon$ in such a way that, provided $|x-1|\lt \epsilon$ we have $|f(x)-2|\lt \frac {\delta}5$. If necessary we take the minumum value of $\epsilon$ and $e$.

Then $$\mid \left(f(x)\right)^2-4\mid=\mid f(x)+2\mid\times\mid f(x)-2\mid\lt 5\times \frac {\delta}5=\delta$$

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  • $\begingroup$ Thanks for your reply. Few questions though: why is it 1 < f(x) < 3 and not 0<f(x)<4, since the limit is 2? Also, how did you get to ${\delta}/5$? Sorry if these are obvious. $\endgroup$
    – iaskdumbstuff
    Oct 22, 2017 at 17:38
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    $\begingroup$ @user494405 I can choose a bound for $f(x)$ so I chose within $1$ unit to avoid any problems with zero or with negative values - I wanted to make sure this was positive and a definite distance away from zero. I just needed something to make the $|f(x)+2|$ factor less than some positive constant (here $5$ works). Likewise I can choose to target $\delta /5$ if I want to, because I can hit any narrow band I choose. $\endgroup$
    – Mark Bennet
    Oct 22, 2017 at 18:23

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