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$f(x)=\displaystyle\lim\limits_{t\rightarrow0}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}$ where [.] denotes the greatest integer function, and $k$ is an integer.

  1. For what values of $k$ will $f(x)$ be continuous $\forall x\in \mathbb R$
  2. For what values of $k$ will $f'(x)$ be continuous $\forall x\in \mathbb R$
  3. For what values of $k$ will $f''(x)$ be continuous $\forall x\in \mathbb R$

$f(x)=\displaystyle\lim\limits_{t\rightarrow0}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\lim\limits_{t\rightarrow0}\frac{\sin(k\pi[x^2-x+\pi])}{\ln (k+[x]^2)}$, Using property $\displaystyle\lim\limits_{x\rightarrow0}\frac{\sin(x)}{x}=1$

$f(x)$ will be a continuous constant function equal to zero $\forall x\in \mathbb R$ if $k>0$ (since log function is not defined for negative numbers or zero). There is a chance that $f(x)$ can be discontinuous at $k=1,x=0$, so we need to check the function at that point.

When $\displaystyle k=1^+,\lim\limits_{t\rightarrow0}\frac{3\sin(k\pi)}{\ln (1^+)}=0/h=0$ where $h$ is an infinitesimally small positive number

Similarly when $\displaystyle k=1^-,\lim\limits_{t\rightarrow0}\frac{3\sin(k\pi)}{\ln (1^-)}=0/h=0$ where $h$ is an infinitesimally small negative number

So $f(x)$ should be continuous and differentiable at at $k=1$, but the answer says $1$ is excluded and $k>1$. What am I missing?

Also can we say the second derivative doesn't exist for a continuous function.

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2 Answers 2

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$f(x)=\displaystyle\lim\limits_{t\rightarrow0}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\lim\limits_{t\rightarrow0}\frac{\sin(k\pi[x^2-x+\pi])}{\ln (k+[x]^2)}$

I think that you also have to consider the case where $t\to \color{red}{0^-}$ for which $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ cannot be used.

For $t\to 0^+$ for which $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$ can be used, it should be $$\displaystyle\lim\limits_{t\rightarrow0^+}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\frac{\sin(k\pi[x^2-x+\pi])}{\ln (k+[x]^2)}$$ not $\displaystyle\color{red}{\lim\limits_{t\rightarrow 0^+}}\frac{\sin(k\pi[x^2-x+\pi])}{\ln (k+[x]^2)}$.

(note that $\displaystyle\lim_{t\to 0^+}\dfrac{\sin(k\pi/e^{1/t})}{k\pi/e^{1/t}}=1$ cannot be used when $k=0$.)

There is a chance that $f(x)$ can be discontinuous at $k=1,x=0$, so we need to check the function at that point.

When $\displaystyle k=1^+,\lim\limits_{t\rightarrow0}\frac{3\sin(k\pi)}{\ln (1^+)}=0/h=0$ where $h$ is an infinitesimally small positive number

If you fix $k=1$, then I think that you cannot take $k=1^+$.


In the following, I'll show my solution.

If $t\to \color{red}{0^+}$, then since $\displaystyle\lim_{t\to 0^+}\dfrac{k\pi}{e^{1/t}}=0$, one has , for $k\not=0$,$$\begin{align}&\displaystyle\lim\limits_{t\rightarrow 0^+}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)} \\\\&=\frac{1}{\ln (k+[x]^2)}\displaystyle\lim_{t\to 0^+}\sin\bigg(\frac{\sin(k\pi/e^{1/t})}{k\pi/e^{1/t}}\cdot k\pi[x^2-x+\pi]\bigg) \\\\&=\frac{1}{\ln (k+[x]^2)}\cdot \sin\bigg(1\cdot k\pi[x^2-x+\pi]\bigg) \\\\&=\frac{0}{\ln (k+[x]^2)}\end{align}$$(For $k=0$, $\displaystyle\lim\limits_{t\rightarrow 0^+}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\frac{0}{\ln ([x]^2)}$.)

If $t\to \color{red}{0^-}$, then $\displaystyle \lim_{t\to 0^-}e^{1/t}=0$ and $$-e^{1/t}\leqslant \sin(k\pi/e^{1/t})e^{1/t}\leqslant e^{1/t}$$ imply $$\lim_{t\to 0^-}\sin(k\pi/e^{1/t})e^{1/t}=0$$so $$\displaystyle\lim\limits_{t\rightarrow 0^-}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\frac{0}{\ln (k+[x]^2)}$$

Therefore, it follows from these that $$f(x)=\displaystyle\lim\limits_{t\rightarrow 0}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}=\frac{0}{\ln (k+[x]^2)}$$

Now, one can say

  • If $k\leqslant 1$, then for $0\leqslant x\lt 1$, $f(x)=\dfrac{0}{\ln(k)}$ is not defined.

  • If $k\gt 1$, then for any $x$, $f(x)=\dfrac{0}{\ln (k+[x]^2)}=0$.

Therefore, one can say

  • $f(x)$ is continuous $\forall x\in \mathbb R$ if and only if $k\gt 1$.

  • $f'(x)$ is continuous $\forall x\in \mathbb R$ if and only if $k\gt 1$.

  • $f''(x)$ is continuous $\forall x\in \mathbb R$ if and only if $k\gt 1$.

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A function $f$ is called continuous at $x=a$ if $$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$$ I think you are missing the crucial $=f(a)$ part.

Also, if $k=1$, you function becomes $$f(x)=\frac{\sin(\pi[x^2-x+\pi])}{\ln (1+[x]^2)}$$ The numerator, being an integer multiple of $\pi$ is always $0$. But, in the limit of $x\to 0^+$, that is, when $x$ is a very small positive real number, the denominator is exactly $(\ln 1)$ as $[x]=0$. So, the function takes $\frac 0 0$ form around $x=0^+$.

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