Aspire Faculty ID #11975 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Let $\mathbb{R}\rightarrow\mathbb{R}$ be any function defined as $f(x)=\begin{cases}{{x}^{\alpha}\sin \frac{1}{{x}^{\beta}}} & {,x\ne0} \\ {0} & {,x=0}\end{cases}$, $\alpha , \beta \in \mathbb{R}$. Which of the following is true? ($\mathbb{R}$ denotes the set of all real numbers)

Solution

The function is $f(x) = x^{\alpha}\sin\left(\dfrac{1}{x^{\beta}}\right)$ for $x \ne 0$, and $f(0)=0$. 
To check continuity at $x=0$, consider: $\displaystyle \lim_{x\to 0} x^{\alpha}\sin\left(\frac{1}{x^{\beta}}\right)$. 
Since $|\sin \theta| \le 1$ for all $\theta$ 
$|x^{\alpha}\sin(1/x^{\beta})| \le |x^{\alpha}|$. 

Now, $\displaystyle \lim_{x\to 0} x^{\alpha} = 0 \quad \text{iff } \alpha > 0$. 
Therefore, $\displaystyle \lim_{x\to 0} x^{\alpha}\sin(1/x^{\beta}) = 0 = f(0)$ iff $\alpha > 0$. 
The value of $\beta$ does not matter because $\sin(1/x^{\beta})$ is always bounded. 
Hence: ${f(x)\text{ is continuous at }x=0\text{ for all }\alpha>0\text{ and }\beta\in\mathbb{R}.}$
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