I need to show that the polynomial $$x^3-x-3$$ is continuous at $x=1$ using epsilon delta proof but I'm facing some problem manipulating the inequality.
Given $$\epsilon>0$$
$$0<|x-1|<\delta$$
Continuity implies $$|f(x)-f(x_0)|<\epsilon$$
so$$x^3-x-6<\epsilon$$
How do I manipulate the last line such that I have something that resembles the $$(x-1)$$?
On
If you set $$ f(x)=x^3-x-3, $$ then we have $$ |f(x)-f(1)|=|x^3-x|=|x(x^2-1)|=|x(x+1)(x-1)|=|x^2+x||x-1| \quad \forall x \in \mathbb{R}. $$ In particular, for $0\le x \le 2$ (i.e. for $|x-1|\le 1$) we have $$ |f(x)-f(1)|\le 6|x-1|. $$ Therefore, given $\varepsilon>0$, if we set $\delta=\frac{1}{6}\min\{1,\varepsilon\}$, then for for every $x \in \mathbb{R}$ satisfying $|x-1|< \delta$ we have $$ |f(x)-f(1)|\le 6|x-1|<6\delta\le \varepsilon. $$ Hence $f(x)=x^3-x-3$ is continuous at $x=1$.
On
in a fairly general context:
1) a constant function is continuous
2) the identity automorphism is continuous
for maps $\mathbb{R} \to \mathbb{R}$ prove by an epsilon-delta argument:
3) the sum/difference of two continuous functions is continuous
4) the product of two continuous functions is continuous
now apply these four propositions to obtain: $x \to 3$ and $x \to x$ are continuous. $x \to x^2$ is continuous. $x \to x^3$ is continuous. $x \to x^3-x$ is continuous. $x \to x^3-x-3$ is continuous
If $|x-1|\lt\delta$ we have $1-\delta\lt x\lt1+\delta$, so we have $0\lt|(x+1)x(x-1)|\lt|(2+\delta)(1+\delta)(\delta)|$, and it is easy to see that the right hand side can be made as small as we like.