How to prove $\displaystyle\lim_{z\to z_0}\dfrac{f(z)}{g(z)}=\dfrac{f'(z_0)}{g'(z_0)}$ How to prove the following


  $f, g$ are differentiable at $z_0$ with $f(z_0) = g(z_0) = 0.$ If $g'(z_0) ≠ 0$ then $\displaystyle\lim_{z\to z_0}\dfrac{f(z)}{g(z)}=\dfrac{f'(z_0)}{g'(z_0)}$
$$\lim_{z\to z_0}\frac{f(z)}{g(z)}=\lim_{z\to z_0}\frac{f(z)-f(z_0)/(z-z_0)}{g(z)-g(z_0)/(z-z_0)}=\frac{\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}}{\lim_{z\to z_0}\frac{g(z)-g(z_0)}{z-z_0}}=\frac{f'(z_0)}{g'(z_0)}$$
i agree with user39483 .its correct .he has taken the right hand side and used definition of derivative .then the denominators cancel out and given values are substituted .hence we get the left hand side ...now i dont know how to write z(naught) in computer so i will use m in place of z(naught)..........................................................................
LHS= lim(z--m) [f(z)/g(z)].....let z=m+h then as z--m then h tends to 0......= lim(h--0)[f(m+h)/g(m+h)]......now using the taylors theorem .....= lim(h--0) [{f(m)+hf'(m)+...+R1} / {g(m)+hg'(m)+...+R2}].....where R1 and R2 are remainders.....since f(m)=g(m)=0 therefore.....= lim(h--0) [ {hf'(m)+terms with higher powers of h} /{hg'(m)+terms with higher power of m}].....= lim(h--0) { h(f'(m)+terms with higher powers of h)} /{h (g'(m)+terms with higher powers of m)} ].....we took h common in previous step and now h cancels out and then we apply  limit.....= f'(m)/g'(m) .....=RHS..... if this solution is incorrect then i would like to know whats wrong with it.