Problems from Discontinuity of functions: Avoidable, Jump and Essential discontinuity

Development:

a) The domain of the function is $Dom(f)=\mathbb{R}--1$ so we must study the continuity of the function in this interval.

The function presents problems exactly at the point $-1$, since it is divided by zero, for what we might think that it might not be continuous.

But let's remember that the definition of continuous function is done on points of the domain, so we cannot apply the concept of continuity or discontinuity at this point.

In the rest of points there are no problems and we always have: $(a\ne -1)$

$${\displaystyle \begin{array}{l}\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a}{lim}{\displaystyle \frac{x^2-1}{x+1}}={\displaystyle \frac{a^2-1}{a+1}}\\ \underset{x\to a^{+}}{lim}f(x)=\underset{x\to a}{lim}{\displaystyle \frac{x^2-1}{x+1}}=\frac{a^2-1}{a+1}\\ f(a)=\frac{a^2-1}{a+1}\end{array}}$$ since the function is continuous.

b) We must study continuity at point $x=1$, since at others it is continuous: $${\displaystyle \begin{array}{l}\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}{\displaystyle \frac{1+x}{1-x}}={\displaystyle \frac{1+1^{-}}{1-1^{-}}}={\displaystyle \frac{2^{-}}{0^{+}}}=+\mathrm{\infty }\\ \underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}{\displaystyle \frac{1+x}{1-x}}={\displaystyle \frac{1+1^{+}}{1-1^{+}}}={\displaystyle \frac{2^{+}}{0^{-}}}=-\mathrm{\infty }\\ f(1)=0\end{array}}$$ Note that we have an essential discontinuity, thus the function is not continuous.

c) The two components of the function are continuous. We only need to check the point $x=3$: $${\displaystyle \begin{array}{l}\underset{x\to 3^{-}}{lim}f(x)=\underset{x\to 3}{lim}x+1=4\\ \underset{x\to 3^{+}}{lim}f(x)=\underset{x\to 3}{lim}x-1=2\\ f(3)=4\end{array}}$$ and since the side limits do not coincide and they are finite we have an inevitable discontinuity.

Solution:

a) Continuous function

b) Function is not continuous. With essential discontinuity at $x=1$.

c) Function is not continuous. With inevitable discontinuity at $x=3$.

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