I'm asked to "prove that each of the following functions is uniformly continuous on the indicated interval".
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$\frac{x^{18}+6x^{12}+12x^6+18}{\left(x-6\right)\left(x-12\right)\left(x-18\right)}$ on (24, 30)
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$\frac{sinx}{x}$ on (0, 1)
Is there a trick to solving these questions?
Or – is it enough to use the theorem that states that if f is continuous on the interval [a,b], then f is uniformly continuous on [a,b]? Would stating that these functions are continuous because their denominator is not equal to 0 in the given interval be enough to prove that they are uniformly continuous?
I've also tried using the definition of uniform continuity but the first question gets way too complicated and I get stuck on the second one.
Any hints or suggestions would be greatly appreciated 🙂
Best Answer
$f(x)=\dfrac{\sin x}{x}$ on $(0,1)$
Thm: A function is said to be uniformly continuous on $(a,b)$ iff it can be extended to a continuous function on $[a,b]$.
Clearly, $f$ is uniformly continuous on $(a,0)$ and $(0,b)$
The continuity at $0$ reflects the differnetiability of $\sin x$ at $0$, which is $\cos0=1$. i.e. $$1=\lim_{x\to0}\dfrac{\sin x - \sin0}{x-0}=\lim_{x\to0}\dfrac{\sin x}{x}$$ and obviously $f(x)$ is continuous at $1$. Therefore we conclude that $f(x)$ can be extended to $[0,1]\implies f(x)$ is continuous on $(0,1)$.
$g(x)=\dfrac{x^{18}+6x^{12}+12x^6+18}{\left(x-6\right)\left(x-12\right)\left(x-18\right)}$ on $(24,30)$
Clearly, $g$ is continuous on the interval without $x=6,12,18$, which implies $g$ is continuous on $[24,30]$, then it's uniformly continuous on $(24,30)$.