A function $f:\Bbb{R}\to\Bbb{R}$ is defined by $f(x)=x$, if $x$ is rational; $\sin(x)$ if $x$ is irrational.
Show that $f$ is differentiable at $0$ and $f'(0)=1$.
Here I'm thinking to apply definition separately for rational and irrational.. and I'm getting same limit... But can we conclude the differentiability with this result??
Yes, the fact that this holds for rationals and irrationals separately (but consistently) does allow you to conclude differentiability. In particular, you are trying to show that the limit $$\lim_{h\rightarrow 0}\frac{f(h)}h=1$$ and you have that $f(x)$ at every point is equal to either $f_1(x)=x$ or $f_2(x)=\sin(x)$ at every point. You have $$\lim_{h\rightarrow 0}\frac{f_1(h)}h=1$$ $$\lim_{h\rightarrow 0}\frac{f_2(h)}h=1$$ Then, you can conclude that the original limit exists as follows: For any $\varepsilon$, you can choose $\delta_1$ and $\delta_2$ such that for any $|x|<\delta_1$ you have $\left|\frac{f_1(h)}h-1\right|<\varepsilon$ and for any $|x|<\delta_2$ you have $\left|\frac{f_2(h)}h-1\right|<\varepsilon$. This is just from definition of the latter two limits. Then, for any $|x|<\min(\delta_1,\delta_2)$ you get that $\left|\frac{f(h)}h-1\right|<\varepsilon$, since one of the two above inequalities will be applicable. Thus, $f(h)$ must have the same derivative as $f_1$ and $f_2$ where those two functions and their derivatives agree.
By definition we need to compute $\displaystyle \lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\displaystyle \lim_{h\to 0}\frac{f(h)-0}{h}=\lim_{h\to 0}\frac{f(h)}{h}$
Notice for any $x\in \mathbb{R}$ we have $|x|\geq|\sin(x)|$ (Easy to check using mean value theorem). That means for $h>0$, we have $\sin(h)\leq f(h)\leq h\Rightarrow \frac{\sin(h)}{h}\leq \frac{f(h)}{h}\leq 1$, by squeeze theorem $\displaystyle \lim_{h\to 0^{+}}\frac{f(h)}{h}=1$. When $h<0$ the inequality reverses and we can also use squeeze theorem to conclude $\displaystyle \lim_{h\to 0^{-}}\frac{f(h)}{h}=1$.
Therefore $\displaystyle \lim_{h\to 0}\frac{f(h)}{h}=1$, the function is differentiable at $0$ and the derivative is 1.
If you put $g(x)=f(x)-x$, then it suffices to show $g'(x)=0$.
If you are able to show that $|g(x)|\le x^2$ then you have $$-x \le \frac{g(x)}x \le x$$ which implies $g'(x)=\lim\limits_{x\to0}\frac{g(x)}x=0$ (using squeeze theorem).
To prove the inequality for $|g(x)|$ you may use some inequality for $\sin x$. Some such inequalities can be found on this site. For example, see this question: Prove that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
