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I have proved that $g$ is continuous on $(0,2)$ and I just wish to check if my solution for $g$ being right continuous at $0$ and hence continuous at $0$ is correct.

$$\lim\limits_{x \to 0^+}g(x) = \lim\limits_{x \to 0^+} \frac{f(0)-f(2x)}{0-2x}= \lim\limits_{x \to 0^+} \frac{f(x)-f(0)}{x-0}=f'(0)$$

Using the fact that $f$ is differentiable, hence $f'_+(0)=f'(0)$

I switched the $2x$ for an $x$ because for small $x$ they're basically the same.

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While your working is more or less correct, I think you could probably be a little more thorough than saying $x$ and $2x$ are "basically the same". Depending on how rigorous you want to be, you might like to use a $\epsilon$-$\delta$ method, but you can also easily use the chain rule. Define $F(x):=f(2x)$. Then $F$ is differentiable and $F'(x)=2f'(2x)$. Then we have $$\lim_{x\to0^+}\frac{f(0)-f(2x)}{0-2x}=\frac12\lim_{x\to0^+}\frac{F(x)-F(0)}x=\frac12F'(0)=f'(0).$$ You can do something similar for the right limit. Define $G(x):=f(2x-2)$. Again, $G$ is differentiable with $G'(x)=2f'(2x-2)$. So $$\lim_{x\to2^-}\frac{f(2x-2)-f(2)}{(2x-2)-2}=\frac12\lim_{x\to2^-}\frac{F(x)-F(2)}{x-2}=\frac12F'(2)=f'(4-2)=f'(2).$$

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Your solution is correct in principle. It could, however, do with some additional justification and/or explanation.

For example, you could say:

Since $a(x) = 0$ and $b(x) = 2x$ for $0 \le x < 1$, it follows that $$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} \frac{f(0)-f(2x)}{0-2x}$$

It depends on how you define $\lim$ whether you can improve on the intuitive leap of swapping $2x$ for $x$. In general, you can prove the following theorem:

Let $h: I \to \Bbb R$ be a continuous function, where $I$ is an open interval, and let $a \in I$. Let $k$ be a function such that $\lim\limits_{x \to h(a)} k(x)$ exists. Then: $$\lim_{x\to a} k(h(x)) = \lim_{x \to h(a)} k(x)$$

where in our present case, $h(x) = 2x$ and $k(x) = \dfrac{f(0)-f(x)}{0-x}$ has a limit at $h(0) = 0$ by virtue of $f$ being differentiable.

However, if you haven't been given a mathematically precise definition of limits (like the $\epsilon$-$\delta$ definition), then proving this type of "obvious" theorem will always feel a little shaky, and you're probably fine justifying this step in your proof in an informal way.

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It looks alright to me but perhaps at this level one should explain why you take $a(x)=0$ and $b(x)=2$ and why is $2x$ and $x$ basically the same when working with limits around 0.

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