Calculus: tangents and limits 
For 1), I know the graph has a tangent but I don't know how to explain.
For 2), is the answer 0? Would anyone mind giving me the steps?
For 3) and 4),  I really have no idea. 
 A: For (2) you can write, $$\frac{\sin(\sin(\theta))}{\sin(\theta)} \cdot \frac{\sin(\theta)}{\theta}$$ Apply the limits individually that will be equal to $1$.
A: For part $4$, because $x$ is approaching from right side of $x$,$f(x)=-x$. And so its derivative is $-1$.
For question $2$, it is the indeterminate form $0/0$, so apply L'Hôpital's rule to get:
$$\cos(\sin(\theta))\cdot \cos(\theta)/1$$
After putting $0$, we get $1\cdot 1=1$.
A: For 1, the graph $g(x)$ near $0$ is helpful:

If a function has a tangent at any point, there must exist the derivative at the point [why?]. So, only if $g'(0)$ exists, then $g(x)$ has a tangent at the origin. The graph shows that such functions really behave oddly near $0$, so its important to deal with them carefully. 
So, computing $g'(0)$ by definition:
$$=\lim_{h\to0}\frac{g(0+h)-g(0)}{h}$$
$$=\lim_{h\to0}\frac{(0+h)\sin(\frac{1}{0+h}))-0}{h}$$ 
This was because $f(0) = 0$ and $f(x)=x\sin x$, which is what would apply when you consider $0+h$.
$$=\lim_{h\to0}\frac{h\sin(\frac{1}{h})}{h}$$
$$=\lim_{h\to0}\sin(\frac{1}{h})$$
which is undefined [why?]. So the derivative does not exists, nor does the tangent. Note that even though $g(x)$ is defined and continuous for every $x$, it fails to differentiable for every $x$. Quite interesting? There is another point to beware of, the very similar function $f(x)=x^2 \sin (\frac{1}{x})$ for $x \ne 0$, and $f(0)=0$, which has a very similar graph, actually does have a derivative at $0$. It reduces to $\lim_{h\to0}h\sin(\frac{1}{h})$, which by squeeze theorem is $0$. These functions are strange, but not at all the strangest, for mathematics has many stranger functions than them.     
For 2, Trafalgar has already given a solution. I will rewrite it here for completeness, skimming the details:
$$=\lim_{\theta\to0}\frac{\sin (\sin \theta)}{\theta}$$
$$=\lim_{\theta\to0}\frac{\sin (\sin \theta)}{\theta} \cdot \frac{\theta}{\sin \theta} \cdot \frac{\sin \theta}{\theta} $$
$$=\lim_{\theta\to0}\frac{\sin (\sin \theta)}{(\sin \theta)} \cdot \frac{\sin \theta}{\theta} $$
$$=\lim_{\theta\to0}\frac{\sin (\sin \theta)}{(\sin \theta)} \cdot \lim_{\theta\to0} \frac{\sin \theta}{\theta} $$
$$=1 \cdot 1 =1$$
For 4, though a graph might be handy, it would better if we do this without it. 
To solve this, we revise that what actually is the absolute value function. It is a function, $|x|$, that for non negative $x$, gives back $x$, and for negative $x$ gives back $-x$. Then your $f(x) = -|x|$  , inverts this and instead for non negative $x$, it gives $-x$. We are only concerned with non negative $x$ (infact only positive $x$), as the limit is approaching $0$ from the right hand side. So, we first compute the derivative for positive $x$: 
$$(\lim_{h\to0} \frac{-|x+h| + x}{h}) = (\lim_{h\to0} \frac{-x-h+x}{h})= -1$$ 
And so the limit, $\lim_{x\to0^+} f'(x)$ would simply be $-1$, as we are only concerned with positive $x$ in this limit.  
