Find the left hand limit of $f(x)$ at $x = 4$ if $f(x) =\frac{|x-4|}{x-4}$ if $x \neq 4$ and $f(x)=0$ if $x=4$ I just began limits for an introductory knowledge of calculus for Physics and I encountered this question in my Mathematics textbook.
So, what I know about limits is that if
$$\lim_{x \rightarrow a^-}f(x) = m$$
It means that as $x \rightarrow a$ from the left side, $f(x) \rightarrow m$ and if it was $x \rightarrow a^+$ instead of $x \rightarrow a^-$, then that would mean that as $x \rightarrow a$ fropm the right side, $f(x) \rightarrow m$.

I also know that if
$$\lim_{x \rightarrow a^-}f(x) = \lim_{x \rightarrow a^+}f(x)$$
It means that
$$\lim_{x \rightarrow a}f(x) \text{ exists and else it does not exist}$$
One last thing that I know about limits is that the statement $x \rightarrow a^-$ can also be written as $x = a-h$, where $h &gt; 0$ and $h \rightarrow 0$ and in the case of $x \rightarrow a^+$, $x = a + h$, $h&gt;0$ and $h \rightarrow 0$
Now, on to the question, it is given that
$$f(x) = \begin{cases} \dfrac{|x-4|}{x-4}&amp;,x \neq 4\\ 0&amp;,x=4 \end{cases}$$
So, as $x \rightarrow 4$d, shouldn't the value of $f(x)$ approach 0, since $f(4)$ is defined and equal to $0$?
Here's how it's done in my book :
$$\lim_{x \rightarrow 4^-}f(x) = \lim_{h \rightarrow 0} f(4-h)$$
$$ = \lim_{h \rightarrow 0} \dfrac{|4-h-4|}{4-h-4} = \lim_{h \rightarrow 0} \dfrac{|-h|}{h}$$
Since $h &gt; 0$, so $-h &lt; 0$ and $|-h| = -(-h) = h$ which means that
$$\lim_{h \rightarrow 0} \dfrac{|-h|}{h} = \lim_{h \rightarrow 0} \dfrac{h}{-h} = \lim_{h \rightarrow 0} -1 = -1$$
I think that I have some misconception either about limits or about functions

Please let me know
Thanks!
When you compute a limit at a certain point, the value of the function at that point does not matter (otherwise, there would be no point computing the limit).
Now,
$$\lim_{x\to4^-}\frac{|x-4|}{x-4}$$ only involves values of $x$ that are less than $4$. But
$$x&lt;4\implies|x-4|=-(x-4)$$ so that
$$x&lt;4\implies\frac{|x-4|}{x-4}=-1.$$
The limit of a constant is that constant.
By a similar reasoning, $\lim_{x\to4^+}\dfrac{|x-4|}{x-4}=1$, and the ordinary limit does not exist.

Intuitively, the limit is the value that you would &quot;naturally&quot; assign to $f$ at $a$ if $f(a)$ was unknown, and it need not be equal to the actual $f(a)$. (In case of a discrepancy, we say that there is a discontinuity.) When we are not able to assign a &quot;natural&quot; value, the limit is said not to exist.
You can start by defining explicitly the function f :
$f(x)  = -1 \quad \forall x &gt; 4 \quad and f(x)  = 1 \quad \forall x &lt; 4$, by the way we can immediatly confirm there is no limit of f on left hand of x = 4.