# $f(x)=\lim\limits_{t\rightarrow0}\frac{\sin(\sin(k\pi/e^{1/t})e^{1/t}[x^2-x+\pi])}{\ln (k+[x]^2)}$ where k is an integer and $x\in \mathbb R$?

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

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

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 $$k\gt 1$$.

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 $$x=0^+$$.