Prove that a function is continuous for every $x \in R$ Prove that the function:
$$
f(x)=\frac{\sqrt{x^2-x+1}}{|\sin(x)-4|-2}
$$
is defined for every $x \in R$ and continuous in every $x \in R$,
So I said that in order for this function to be defined we demand that:
$x^2-x+1 > 0$ which is always true
and
$ |\sin(x)-4|-2 \ne 0 $ which is also true always,
But how do I prove that a function is continuous? How I can use what I found to show that?
Thanks
 A: The function $x\mapsto\sqrt{x}$ is continuous; the function $x\mapsto x^2-x+1$ is continuous and only takes on positive values. Therefore $x\mapsto\sqrt{x^2-x+1}$ is continuous.
Since $\sin x\le 4$, you have $|\sin x-4|=4-\sin x$, so the denominator is $2-\sin x$; the function $x\mapsto 2-\sin x$ is continuous and only takes on positive values.
The quotient of continuous functions is defined and continuous wherever the denominator is nonzero.
A: $sin$ has values between 1 and -1, so that the numerator is always bigger than 1 and is never zero. The numerator is well defined as you point out. As the composition, sum, product and inverse (for the multiplication) of continuous functions is a continuous function, your function is continuous.
A: If $f(x)$ is non-negative and continuous, so it is $\sqrt{f(x)}$. If $f(x)$ and $g(x)$ are continuous and $g(x)$ is bounded away from zero, then $\frac{f(x)}{g(x)}$ is continuous. $x^2-x+1$ is non-negative since its discriminant is negative. $|\sin x-4|$ is bounded away from $2$ since $\sin x-4$ belongs to $[-5,-3]$ so its absolute value belongs to $[3,5]$. Now, just put all together.
A: To show continuity you can either take the definition and do $\varepsilon$/$\delta$ calculations. For this function that looks like something that might be quite hard (I haven't bothered to try).
The alternative is to use results about the continity of composited functions. The only tricky part is that the ratio of two continous functions is only continuos if the denominator is not zero, but you have already proven that.
