# check for Continuity $x \rightarrow 2x^4-2$

Given the function: $f:\Bbb{R}\rightarrow \Bbb{R},\quad x \rightarrow 2x^4-2$

i am asked to check for Continuity for all values of $x$. i am now overasked how to do this since $\Bbb{R}$ is not a small range of numbers.

Definition: the Continuity of a function means that a function isnot to have any holes o jumps in its graph.

how do i check this for all domain values of $\Bbb{R}$?

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What have you tried so far? Do you know about the $\epsilon - \delta$ criteria ? – k1next Dec 23 '12 at 7:12
Okay the problem is, that the real definition about continuity is not "does not have any jumps or holes", but rather that if you take two values on the x-axis that are arbitary close to each other, then the function values on the y-axis also become arbitary close. – k1next Dec 23 '12 at 7:20
If you've learned the limit laws, you can compute $$f\left( \lim_{x \to a} x \right)$$ and $$\lim_{x \to a} f(x)$$ to see if they're equal. – Hurkyl Dec 23 '12 at 10:07

## 1 Answer

That's your definition for continuity? Well I suggest looking the graph of $f$ and see if there are any "holes" or "jumps".

Let's do this in a rigorous manner:

The $\epsilon-\delta$ definition: Let $a\in X\subseteq \mathbb{R}$ and $f:X\to \mathbb{R}$. Then we say that $f$ is continuous at $a$ if $$\forall \epsilon>0\exists \delta>0:\ \left|x-a\right|<\delta\implies \left|f(x)-f(a)\right|<\epsilon$$ If $f$ is continuous at every point of $A\subseteq X$ then $f$ is said to be (pointwise) continuous on $A$.

It is usually hard to prove continuity using the $\epsilon-\delta$ definition. When one has studied limits of sequence this becomes much simpler using the following equivalent definition due to Heine :

Let $a\in X\subseteq \mathbb{R}$ and $f:X\to \mathbb{R}$. Then $f$ is continuous at $a$ iff for every sequence $(x_n)$ of elements of $X$, $x_n\to a\implies f(x_n)\to f(a)$

Using this we have to prove that if $x_n\to a$ for arbitrary $a\in \mathbb{R}$ then $f(x_n)\to f(a)$ or equivalently $2x_n^4-2\to 2a-2$. I will let you prove that.

Sidenote:

Even as is, your non rigorous definition is wrong because it allows essential discontinuities to be continuities. Look at the graph of $f(x)=\sin (1/x)$ $x\neq 0$ and $f(0)=0$. Is $f$ continuous?

EDIT: We need to prove that $$x_n\to a\implies f(x_n)\to f(a)$$ not for a particular $x_n,a$ but for all sequences $x_n$ and points $a$.

Now let $x_n$ be a sequence and $a\in \mathbb{R}$ so that $x_n\to a$. Then $f(x_n)=2x_n^4-2$ while $f(a)=2a^4-2$. We need to prove that $$f(x_n)\to f(a)\iff 2x_n^4-2\to 2a^4-2\iff x_n^4\to a^4$$ (the equivalencies are due to the limit laws). So the problem moves to proving that $x_n^4\to a^4$. But this also follows from the limit laws as $x_n\to a\implies x_n\cdot x_n\to a\cdot a\implies...\implies x_n^4\to a^4$

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Okay, thank you, i will try to prove then – doniyor Dec 23 '12 at 7:22
@doniyor No! You mustn't show that $x_n\to a$. You must prove that $f(x_n)\to f(a)$ assuming $x_n\to a$ – Nameless Dec 23 '12 at 8:04
Nameless, i got it finally. i took $x_{n}:=\frac{2n+1}{1+n}$ and $x_{n} \rightarrow 2$ so now i have to show. $2(\frac{2n+1}{1+n})^4-2 \rightarrow 2(2)^4-2$, right? – doniyor Dec 23 '12 at 8:33
@doniyor That is just one special sequence $x_n$ that convergese to just one special real number $2$. You have to show the same statement for all sequences converging to any real number. – Hagen von Eitzen Dec 23 '12 at 9:52
@HagenvonEitzen thanks, but how do i show it? thru induction? – doniyor Dec 23 '12 at 9:59