Let f(x)=_0( x-x^2/(x-1)1+x^2/(x-1)), xR.

Question

Let $f(x)=\displaystyle\lim_{	heta	o0}\left(\frac{\cos\pi x-x^{2/	heta}\sin(x-1)}{1+x^{2/	heta}(x-1)}ight),\ x\in\mathbb{R}$.

Consider the two statements:

(I) $f(x)$ is discontinuous at $x=1$.

(II) $f(x)$ is continuous at $x=-1$.

Accepted Answer

Let $n=2/\theta\to\infty$ as $\theta\to0$ (through even integer values, as required by the working below). For $|x|<1$: $x^n\to0$, so $f(x)=\cos(\pi x)$. For $|x|>1$: dividing numerator and denominator by $x^n$, $$f(x)=\lim_{n\to\infty}\frac{\dfrac{\cos\pi x}{x^n}-\sin(x-1)}{\dfrac{1}{x^n}+(x-1)}=\frac{-\sin(x-1)}{x-1}$$ At $x=1$: substituting directly ($x^n=1$ regardless of $n$): $f(1)=\dfrac{\cos\pi-0}{1+0}=-1$. At $x=-1$: since $n$ is even, $x^n=1$; $f(-1)=\dfrac{-1-\sin(-2)}{1+(-2)}=\dfrac{-1+\sin2}{-1}=1-\sin2$. So $$f(x)=\begin{cases}\dfrac{-\sin(x-1)}{x-1}, & x\in(-\infty,-1)\cup(1,\infty)\ \cos(\pi x), & x\in(-1,1)\ -1, & x=1\ 1-\sin2, & x=-1\end{cases}$$ Checking Statement (I) at $x=1$: $f(1)=-1$. LHL $=\lim_{x\to1^-}\cos(\pi x)=\cos\pi=-1$. RHL: with $u=x-1\to0^+$, $\lim\dfrac{-\sin u}{u}=-1$. Since LHL $=$ RHL $=f(1)=-1$, $f$ is continuous at $x=1$. So Statement (I) is False. Checking Statement (II) at $x=-1$: $f(-1)=1-\sin2$. LHL ($x\to-1^-$, $|x|>1$ region): $\dfrac{-\sin(-1-1)}{-1-1}=\dfrac{\sin2}{-2}=-\dfrac{\sin2}{2}$. RHL ($x\to-1^+$, $|x|<1$ region): $\cos(-\pi)=-1$. Since $-\dfrac{\sin2}{2}\ne-1$, the two one-sided limits disagree, so $f$ is discontinuous at $x=-1$. So Statement (II) is False. Hence, neither (I) nor (II) is true.

Solutions · 1