Discuss the continuity of the function f, where f is defined by
$f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$
The given function $f$ is $f(x)=\left\{\begin{array}{l}-2, \text { if } x \leq-1 \\ 2 x, \text { if }-11\end{array}\right.$
The given function is defined at all points of the real line.
Let c be a point on the real line.
Case I:
If $c<-1$, then $f(c)=-2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(-2)=-2$
$\therefore \lim _{x \rightarrow c} f(x)=f(c)$
Therefore, $f$ is continuous at all points $x$, such that $x<-1$
Case II:
The left hand limit of f at x = −1 is,
$\lim _{x \rightarrow-1} f(x)=\lim _{x \rightarrow-1^{-}}(-2)=-2$
The right hand limit of f at x = −1 is,
$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}(2 x)=2 \times(-1)=-2$
$\therefore \lim _{x \rightarrow-1} f(x)=f(-1)$
Therefore, f is continuous at x = −1
Case III:
If $-1
$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x)=2 c$
$\therefore \lim _{x \rightarrow c} f(x)=f(c)$
Therefore, f is continuous at all points of the interval (−1, 1).
Case IV:
If $c=1$, then $f(c)=f(1)=2 \times 1=2$
The left hand limit of f at x = 1 is,
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}(2 x)=2 \times 1=2$\
The right hand limit of f at x = 1 is,
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} 2=2$
$\therefore \lim _{x \rightarrow 1} f(x)=f(c)$
Therefore, f is continuous at x = 2
Case V:
If $c>1$, then $f(c)=2$ and $\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2)=2$
$\lim _{x \rightarrow c} f(x)=f(c)$
Therefore, $f$ is continuous at all points $x$, such that $x>1$
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.
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