Continuity of piecewise function involving defined by $f(x) = x\lfloor \frac{1}{x} \rfloor$ if $x \neq 0$ and $f(x) = 1$ when $x = 0$

Question

Draw the graph and study the continuity of the function $$f(x)=\begin{cases} x\lfloor \frac1x \rfloor, & x \ne 0 \\ 1, &x=0 \end{cases} $$

Any help with how to solve something like that.

I am self studying calculus to prepare for engineering and while I can deal with normal piecewise functions for some reason this one confuses me.

Answer 1

You need to study $$f(x)=x\bigg\lfloor{\frac{1}{x}}\bigg\rfloor$$

We know that $\lfloor{x}\rfloor$ is discontinuous at integers, therefore $\bigg\lfloor{\frac{1}{x}}\bigg\rfloor$ i.e $f(x)$ is dicontinuous at $x=\frac{1}{z}$ where $z$ is an integer.

Now at $x=0$,

$$\lim_{x\to{0}}f(x)=\lim_{x\to{0}}x\bigg(\frac{1}{x}-\bigg\{\frac{1}{x}\bigg\}\bigg)$$

$$\lim_{x\to{0}}f(x)=\lim_{x\to{0}}1-x\bigg\{\frac{1}{x}\bigg\}$$

$$\lim_{x\to{0}}f(x)=1$$

Therefore its continuous at $x=0$.

Here's the graph for the better picture,

Answer 2

Discern the following cases:

• $x>1$ then $f(x)=x\lfloor\frac1x\rfloor=x\cdot0=0$
• $\frac1{n+1}<x\leq\frac1{n}$ for $n=1,2,\dots$ then $f(x)=x\lfloor\frac1x\rfloor=x\cdot n\in\left(\frac{n}{n+1},1\right]$
• $x=0$ then $f(x)=1$
• $-\frac1{n-1}<x\leq-\frac1n$ for $n=2,3,\dots $ then $f(x)=x\lfloor\frac1x\rfloor=x\cdot(-n)\in\left[1,\frac{n}{n-1}\right)$

Answer 3

write [1/x] as 1/x - {1/x} so you have x( 1/x - {1/x}) 1- x {1/x} x{1/x} limit is 0 as it very small multiplied by bounded value. hence limit is 1-0= 1 hence contineous

i have discussed at 0.

Answer 4

Is this what the graph of $f(x)$ looks like?

(This should be a comment, but contains a graphic)