Q.If $f(x)=\begin{cases}x+1, & x<2\\ 3x-1, & x\ge2\end{cases}$, find $\lim_{x\to2}f(x)$ if it exists.

For $x<2$, $f(x)=x+1$, so $\lim_{x\to2^-}f(x)=\lim_{x\to2^-}(x+1)=2+1=3$. For $x\ge2$, $f(x)=3x-1$, so $\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x-1)=3(2)-1=5$. Since the left-hand limit $3$ and the right-hand limit $5$ are unequal, $\lim_{x\to2}f(x)$ does not exist -- even …

Substitute $x=2$ into each piece's own formula ($x+1$ for the left-hand limit, $3x-1$ for the right-hand limit) since each piece is a polynomial (direct subst …

Using the wrong piece for one side (e.g. substituting into $3x-1$ for the left-hand limit); concluding the limit …