Problems from Side continuity

Development:

a) The functions that define $f(x)$ are continuous, so the only point where we could have some problem is at $x=1$, where the two subfunctions meet: $${\displaystyle \begin{array}{l}\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1}{lim}3x=3\\ \underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1}{lim}(2x+1)=2+1=3\\ f(1)=3\end{array}}$$ and since the side limits coincide with the value of the function, the function is continuous.

b) The function ${\displaystyle \frac{1}{x}}$ is continuous in its domain. We need to verify if $f(x)$ is continuous at zero: $${\displaystyle \begin{array}{l}\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}{\displaystyle \frac{1}{x}}=+\mathrm{\infty }\\ \underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}{\displaystyle \frac{1}{x}}=-\mathrm{\infty }\\ f(0)=0\end{array}}$$ Therefore the limits do not coincide with the function at the zero; the function is not continuous.

c) The functions that define $f(x)$ are continuous so we only need to verify the points $x=1$ and $x=-1$, where the different subfunctions meet:

Continuity at $x=1$: $${\displaystyle \begin{array}{l}\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1}{lim}2x=2\\ \underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1}{lim}(x^2+1)=1+1=2\\ f(1)=2\end{array}}$$ therefore the function is continuous at $x=1$.

Continuity at $x=-1$: $${\displaystyle \begin{array}{l}\underset{x\to -1^{+}}{lim}f(x)=\underset{x\to -1}{lim}(x^2+1)=(-1{)}^2+1=2\\ \underset{x\to -1^{-}}{lim}f(x)=\underset{x\to -1}{lim}2x=-2\\ f(-1)=-2\end{array}}$$ therefore the function is not continuous at $x=-1$, so we will not have a continuous function.

Solution:

a) Continuous function

b) Discontinuous function

c) Discontinuous function

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