Show that $ \lim_{x\to 1} f(x)$ exist Given a function $f:[0,2]\to\Bbb R$:
$$f(x)=
\begin {cases}
4x-3    & 0 \le x \le 1 \\
3x^2-2x & 1 \le x \le 2
\end {cases}$$
Show that $\displaystyle \lim_{x\to 1} f(x)$ exists.
My Attempt
$$\begin{array}{rcl}
\textrm {L.H.L} &=& \displaystyle \lim_{x\to 1^-} f(x)\\
                &=& \displaystyle \lim_{x\to 1^-} (4x-3)
\end{array}$$
How do I proceed further?
 A: Note that $4x-3 \to 1$ as $x \to 1-$. You only have to check the limit of $f(x)$ as $x \to 1+$. By construction we have $f(x) = 3x^{2}-2x$ for all $1 < x \leq 2$; and $3x^{2}-2x \to 1$ as $x \to 1+$. So $\lim_{x \to 1}f(x) = 1$.
A: 
How do I proceed further?

Don't stop after the left-handed limit, also find the right-handed limit and verify that they are the same. If:
$$f(x)=\begin{cases}
\color{blue}{4x-3} &  {0\le x\leq 1}\\ 
\color{red}{3x^2-2x} &  {1\le x\le 2}
\end {cases}$$
then:
$$\lim_{x\to 1^{-}} f(x) = \lim_{x\to 1^{-}} \left( \color{blue}{4x-3} \right) = \ldots \quad \mbox{and} \quad 
\lim_{x\to 1^{+}} f(x) = \lim_{x\to 1^{+}} \left( \color{red}{3x^2-2x} \right) = \ldots $$
Both functions are polynomials and thus continuous everywhere: you find the limits by simply plugging in. Can you fill the gaps/dots?

Strictly speaking, the way the function is defined on the two intervals implies in a way that the limit exists because having the double $\color{green}{\le}$ at the end point $x=1$ can only be meaningful if the function values for both expressions agree (since a function can't have more than one function value at any point):
$$f(x)=\begin{cases}
{4x-3} &  {0\le x\color{green}\le 1}\\ 
{3x^2-2x} &  {1\color{green}\le x\le 2}
\end {cases}$$
A: The given function pieces are continous, as they are polynomials.
Then
$$\lim_{x\to1^-}f(x)=\lim_{x\to1^-}4x-3=4\cdot1-3=1$$ and
$$\lim_{x\to1^+}f(x)=\lim_{x\to1^+}3x^2-2x=3\cdot1-2\cdot1=1.$$
As they are equal, the ordinary limit exists.
