# Why are we solving the limits in this way?

The question is-

Find $\lim_{x\to 5} f(x)$ if it exists $f(x)=\frac{x^2-9x+20}{x-[x]}$ where [.] is G.I.F.

Now, my teacher solved it like this-

$\lim_{x\to 5^+} \frac{(x-5)(x-4)}{(x-5)}$

Now, (x-5) gets canceled and gives us

$x-4$ $=$ $5-4=1$

Similarly, for $5^-$, we get the answer as 0.

Here is where my question comes-

I thought we can put the value of limit only simultaneously and not individually as my teacher has done in [x]. Is it possible to insert values of limit separately?

• G.I.F.? Don't know what that means. – Paul Evans May 29 '17 at 10:14
• @PaulEvans I suspect "Greatest Integer Floor"? – Brondahl May 29 '17 at 10:55
• @PaulEvans It's the "Greatest Integer Function", which agrees with the "floor function". – Mark S. May 29 '17 at 13:09

What is happening is that for values of $x > 5$ that are close to $5$, $\lfloor x \rfloor = 5$. And, similarly, for values of $x < 5$ that are close to $5$, $\lfloor x \rfloor = 4$.
So, it is not the case the $\lim_{x \to 5+} \lfloor x\rfloor$ or $\lim_{x \to 5^-} \lfloor x \rfloor$ is taken separately; it is the case that $\lfloor x\rfloor$ is constant for relevant $x$.
Note that $x=[x]+\{x\}$, then: $\lim_\limits{x\to 5} \frac{(x-5)(x-4)}{x-[x]}=\lim_\limits{x\to 5} \frac{(x-5)(x-4)}{\{x\}}.$ Thus:
$$\lim_\limits{x\to 5^+} \frac{(x-5)(x-4)}{\{x\}}=\lim_\limits{x\to 5^+} \frac{\require{cancel} \cancel{\{x\}}(x-4)}{\cancel{\{x\}}}=1.$$ $$\lim_\limits{x\to 5^-} \frac{(x-5)(x-4)}{\{x\}}=\lim_\limits{x\to 5^-} \frac{0\cdot1}{1}=0.$$