Let f(x)= 0lim( x-x^((2)/( )) (x-1)1+x^((2)/( ))(x-1)),x R.

Question

Let $f(x)=\underset{	heta 	o 0}{lim}(\frac{\cos \pi x-{x}^{(\frac{2}{	heta })}\sin (x-1)}{1+{x}^{(\frac{2}{	heta })}(x-1)}),x\in R$. Consider the following two statements :
(I) $f(x)$ is discontinuous at $x=1$.
(II) $f(x)$ is continuous at $x=-1$.
Then,

Accepted Answer

The function is defined as :

$f(x)=\underset{\theta \to 0}{lim}(\frac{\cos \pi x-{x}^{(\frac{2}{\theta })}\sin (x-1)}{1+{x}^{(\frac{2}{\theta })}(x-1)})$

Let $n=\frac{2}{\theta }$. As $\theta \to 0,n\to \infty$ (specifically, let's consider $\theta $ approaching 0 such that the power grows). The behavior of the function depends on the value of $x$ because of the term ${x}^{n}$.

Case 1 : $|x|

$f(x)=\frac{\cos \pi x-0}{1+0}=\cos \pi x$

Case 2 : $|x|>1$ As $n\to \infty,{x}\^{n}\to \infty$. We divide the numerator and denominator by ${x}\^{n}$ :

$f(x)=\underset{n\to \infty}{lim}\frac{\frac{\cos \pi x}{{x}\^{n}}-\sin (x-1)}{\frac{1}{{x}\^{n}}+(x-1)}=\frac{0-\sin (x-1)}{0+(x-1)}=\frac{-\sin (x-1)}{x-1}$

Case 3 : $x=1$ If $x=1$, then $\cos (\pi )=-1$ and $\sin (0)=0$.

$f(1)=\frac{-1-({1}\^{n}\cdot 0)}{1+({1}\^{n}\cdot 0)}=-1$

Case 4 : $x=-1$ If $x=-1$, then $\cos (-\pi )=-1$ and $\sin (-2)=-\sin (2)$. Since $n=2/\theta $ is even, $(-1{)}\^{n}=1$.

$f(-1)=\frac{-1-(1\cdot \sin (-2))}{1+(1\cdot -2)}=\frac{-1+\sin 2}{-1}=1-\sin 2$

So $f(x)$ become

$f(x)=$$\begin{gathered} \frac{-\sin (x-1)}{x-1} x\in (-\infty,-1)\cup (1,\infty) \ \cos \pi x x\in (-1,1) \ 0 x=1 \ 1-\sin 2 x=-1 \end{gathered}$$$

Analyze the Statements

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

Value at $x=1$ : $f(1)=-1$.

Left-hand limit ( $x\to {1}\^{-}$) : Uses the $|x|
Right-hand limit $(x\to {1}^{+})$ : Uses the $|x|>1$ definition: $\underset{x\to {1}^{+}}{lim}\frac{-\sin (x-1)}{x-1}$. Using the standard limit $\underset{u\to 0}{lim}\frac{\sin u}{u}=1$, this equals -1. Since $LHL=RHL=f(1)$, the function is continuous at $x=1$.

Conclusion : Statement (I) is False.

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

Value at $x=-1$ : $f(-1)=1-\sin 2$.

Left-hand limit $(x\to -{1}^{-})$: Uses $|x|>1$ definition : $\frac{-\sin (-1-1)}{-1-1}=\frac{-\sin (-2)}{-2}=\frac{\sin 2}{-2}$.

Right-hand limit ( $x\to -{1}^{+}$

Solutions · 1