How I find $\beta$ for to do this function continuous?

Question

My function is: $f(x) = x^3 $ if $0\leq x \leq 1 $ or $x^2-\beta x+ \beta$ if $1 < x \leq 2$ .

My answer:

For this function to be continuous the $\lim_{x\to1^+} x^2 - \beta x + \beta = \lim_{x\to1^-} x^3$, when a apply the limit I have this:

\begin{align} \lim_{x\to1^+} x^2 - \beta x + \beta = 1 - \beta +\beta\\ \end{align}

\begin{align} \lim_{x\to1^-} x^3 = 1 \end{align}

I want a value for $\beta$ but he is canceling there.

Answer 1

I see nothing wrong. Your function is continuous for all $\beta\in\mathbb{R}$.

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Now, if you're interested in which $\beta$, if any, that will make $f$ differentiable on $(0,1)$, then you just repeat the analysis but with the difference ratio.

Note: $$ \lim_{h\downarrow 0}\frac{f(1+h)-f(1)}{h}=\lim_{h\downarrow 0}\frac{(1+h)^2-\beta(1+h)+\beta-1}{h}=\lim_{h\downarrow 0}(2+h-\beta)=2-\beta, $$ and $$ \lim_{h\uparrow 0}\frac{f(1+h)-f(1)}{h}=\lim_{h\uparrow 0}\frac{(1+h)^3-1}{h}=3. $$ What can $\beta$ be? In the end, your $f$ should look something like this:

Answer 2

A function $f$ is called continuous in $x=1$ if $\lim\limits_{x\uparrow1}f(x)=\lim\limits_{\downarrow1}f(x)=f(1)$. By your definition of $f$, it follows easily that $f(1)=1^3=\lim\limits_{x\uparrow1}f(x)$ and that for all $\beta\in\mathbb{C}$, it follows that $\lim\limits_{x\downarrow1}f(x)=1$, so $f$ is continuous in $x=1$ for all values of $\beta$.