Problems from Continuity of a function at a point

Let's look at the continuity of the function

$$\displaystyle f(x)=\left\{\begin{array}{rcl} x-3 & \mbox{ if } & x\neq3 \\ 0 & \mbox{ if } & x=3 \end{array}\right.$$

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Development:

The functions that define $$f(x)$$ are continuous since they are polynomial, therefore they can only be discontinuous if the two functions do not connect when $$x=3$$.

$$$\displaystyle \begin{array} {l} \lim_{x \to 3^-}f(x)=\lim_{x \to 3} (x-3)= 0 \\ \lim_{x \to 3^+}f(x)=\lim_{x \to 3} (x-3)= 0 \\ f(3)=0 \end{array}$$$ As the lateral limits coincide with the value of the function, the function is continuous.

Solution:

The function is continuous when $$x=3$$ and in all $$\mathbb{R}$$.

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Let's look at the continuity of the function

$$\displaystyle f(x)=\left\{\begin{array}{rcl} x^2+2 & \mbox{ if } & x<1 \\ 3x & \mbox{ if } & x \geq 1\end{array}\right.$$

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Development:

The functions that define $$f(x)$$ are continuous since they are polynomial, therefore they can only be discontinuous if they do not connect when $$x=1$$.

$$$\displaystyle \begin{array} {l} \lim_{x \to 1^-}f(x)=\lim_{x \to 1} (x^2+2)= 1^2+2=3 \\ \lim_{x \to 1^+}f(x)=\lim_{x \to 1} (3x)= 3 \\ f(1)=3\cdot1=3 \end{array}$$$ As the lateral limits coincide with the value of the function, the function is continuous.

Solution:

The function is continuous when $$x=1$$ and in all $$\mathbb{R}$$.

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