Question
Find $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$, where $\text{f}(\text{x})=\begin{cases}\text{x}^2-1,& x \leq 1\\-\text{x}^2-1, & x > 1\end{cases}$

Answer

The given function is$\text{f}(\text{x})=\begin{cases}\text{x}^2-1,& x \leq 1\\-\text{x}^2-1, & x > 1\end{cases}$
$\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})=\lim\limits_{\text{x}\rightarrow1}[\text{x}^2-1]=1^2-1=1-1=0$ $\lim\limits_{\text{x}\rightarrow1^+}\text{f}(-\text{x})=\lim\limits_{\text{x}\rightarrow1}[\text{x}^2-1]=-1^2-1=-1-1=-2$ It is observed that $\lim\limits_{\text{x}\rightarrow1^-}\text{f}(\text{x})\neq\lim\limits_{\text{x}\rightarrow1^+}\text{f}(\text{x})$. Hence, $\lim\limits_{\text{x}\rightarrow1}\text{f}(\text{x})$ does not exist.

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