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$f(x)=[x]+<x>^{0.5}$ Is this continuous and differentiable at $x=1$?

Where $[x]$ is the greatest integer function and $<x>$ represents the integer part of x.

Please help with this.

Is it discontinuous because when $x=1-\varepsilon $ , $f(x)=0+\sqrt{\varepsilon}$

and when $x=1+\varepsilon $ , $f(x)=1+\sqrt{\varepsilon}$

onelessproblem
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  • Both the answers you have received take $\lt x \gt$ as the fractional part of $x$. Your question says it is the integer part, which would match $[x]$ as you have defined it. If you meant $\lt x \gt$ to be the fractional part, please edit your post. Your computation at $x=1+\varepsilon$ uses the fractional part. Your computation at $x=1-\varepsilon$ does not use either one correctly. – Ross Millikan May 05 '17 at 02:08

2 Answers2

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The fractional part of $x$ is $\langle x\rangle=x-[x]$, so when $0<x<1$ then the integer part of $x$ is zero, while the fractional part of $x$ is $\langle x\rangle=x-[x]=x-0$. Hence $f(x)=0+\sqrt{x}$ for $0<x<1$. On the other hand, when $1\le x<2$ then the integer part of $x$ is $1$ and so fractional part of $x$ is $\langle x\rangle=x-[x]=x-1$. Hence, $f(x)=1+\sqrt{x-1}$. I'll leave it to you to check continuity and differentiability.

Gio67
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For $\,0 \le x < 1$,

$$f(x) = \sqrt{x}$$

For $1 \le x < 2$,

$$f(x) = 1 + \sqrt{x-1}$$

It follows that $f$ is continuous at $x=1$, since \begin{align*} &\bullet\;f(1) = 1\\[4pt] &\bullet\;\lim_{x \to 1^{-}}f(x) = \lim_{x \to 1^{-}}\sqrt{x} = 1\\[4pt] &\bullet\;\lim_{x \to 1^{+}}f(x) = \lim_{x \to 1^{+}}1+\sqrt{x-1} = 1 \end{align*}

But $f$ is not differentiable at $x=1$, since \begin{align*} \lim_{h \to 0^{+}} \frac{f(1+h)-f(1)}{h} &= \lim_{h \to 0^{+}} \frac{\left(1 + \sqrt{h}\right)-1}{h}\\[4pt] &= \lim_{h \to 0^{+}} \frac{\sqrt{h}}{h}\\[4pt] &= \lim_{h \to 0^{+}} \frac{1}{\sqrt{h}}\\[4pt] &=\infty \end{align*}

quasi
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