The function f x 1,&|x| 1 1 n^2 ,&1 n <|x|<1 n-1 ,n=2,3,... 1. Is… - Vidyadip

Question

The function $\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$

Is discontinuous at finitely many points.
Is continuous everywher.
Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0
None of these.

✓

Answer

Is discontinuous only at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

Solution:
Given function is
$\text{f{x}}\begin{cases}1,&|\text{x}|\geq1\\\frac{1}{\text{n}^2},&\frac{1}{\text{n}}<|\text{x}|<\frac{1}{\text{n}-1},\text{n}=2,3,...\end{cases}$
Consider,
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^-}\text{f(x)}=\frac{1}{\text{n}^2}$
$\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}^+}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{1}{\text{n}}}1=1$
Hence, fnuction is discontinuous at $\text{x}=\pm\frac{1}{\text{n}},\text{n }\in\text{ z}-\{0\}$ and x = 0

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