Given the function: $f(x)=xe^x+\sin(θ), x\le0$ and $f(x)=x\ln x+θ, x>0$ which is continuous.
I) Show that $θ=0.$
II) Show that $f$ is not differentiable at $x_0=0.$
Personal work:
I) Because $f$ is continuous (hypothesis) then it applies: $$\lim\limits_{x\to 0^-}{f(x)}=\lim\limits_{x\to 0^+}{f(x)}=f(0). (1)$$ •$\lim\limits_{x\to 0^-}{f(x)}=\lim\limits_{x\to 0^-}(xe^x+\sin(θ))=\dots=\sin(θ)$
•$\lim\limits_{x\to 0^+}{f(x)}=\lim\limits_{x\to 0^+}{(x\ln x+θ)}=\dots=θ$
•$f(0)=0*e^0+\sin(θ)=\cdots=\sin(θ)$
So, (1) is equals to:
$$\sin(θ)=θ=\sin(θ)\iff \sin(θ)=θ-\sin(θ)\iff 2\sin(θ)=θ \iff \sin(θ)={θ\over 2}$$
The problem? How can I correctly compute $\sin(θ)={θ \over 2}$
First observe that since $x e^x + sin(\theta)$ is continuous and defines f on $x \geq 0$, you only need to worry about $\lim_{x \to 0^-} f(x) = f(0)$.
This gives you $sin(\theta) = \theta$ as you correctly computed, then observe this implies $\theta=0$. In particular, the only intersection of the curves $\sin t$ and $t$ is at the origin. (Formally you could argue that any solution must be in $[-1,1]$, and use a calculus argument to demonstrate that no other roots exist).