Aspire Faculty ID #16694 · Topic: CUET 2023 · Just now
CUET 2023

The point(s) at which the function $f$ given by $f(x)=\begin{cases} \dfrac{x}{|x|}, & x<0 \\ -1, & x\ge 0 \end{cases}$ is continuous is/are.

Solution

For $x<0$, $|x|=-x$ so $f(x)=\dfrac{x}{-x}=-1$ (constant) ⇒ continuous. For $x>0$, $f(x)=-1$ ⇒ continuous. At $x=0$: $\displaystyle \lim_{x\to0^-} f(x)=-1,\quad \lim_{x\to0^+} f(x)=-1,\quad f(0)=-1$ Hence $f$ is continuous at $x=0$. Therefore, $f$ is continuous for all real $x$.
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