calculus, finding continuous and differential of function

Question

$$\forall a\in \mathbb{R}, f(x)= \begin{cases}x+xe^{1/x} & \text{ if } x<0 \\ 0 & \text{ if } x=0 \\ {{(a-2\cos x)}/{sin x}}& \text{ if } x>0 \end{cases}$$

1. find all $a$ values for $f(x)$ will be continuous at $x=0$
2. find all $a$ values for $f(x)$ will be differentiable at $x=0$

Answer 1

f is continuous at 0 if $\lim_{x\to 0^-}f(x) =0=\lim_{x\to 0^+}f(x)$ that is $$0= \lim_{x\to 0^-}(x+xe^{1/x})=\lim_{x\to 0^+} \frac{(a-2\cos x)}{\sin x} = \lim_{x\to 0^+} \frac{(a-2\cos x)}{ x} \lim_{x\to 0^+}\frac{x}{\sin x}$$

But we have , $$ \lim_{x\to 0^-}(x+xe^{1/x}) = 0~~~and ~~~\lim_{x\to 0^+}\frac{x}{\sin x}=1$$ sO, it is true if and only if

$$\lim_{x\to 0^+} \frac{(a-2\cos x)}{ x} =0 =\lim_{x\to 0^+} \frac{(\frac{a}{2}-1+ 1-\cos x)}{ x} = \lim_{x\to 0^+} \frac{\frac{a}{2}-1}{ x}$$

Since $$\lim_{x\to 0^+} \frac{1-\cos x}{ x} =\cos'(0)=\sin (0)= 0$$ Hence the only possibility is $$\color{red}{a=2}$$