Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$
So
$$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$
Then
$$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$
That's
$$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$
Finally
$$\frac{3}{(1+\sqrt[3]{4-3x})}$$
But this evaluates to
$$\frac{3}{2}$$
When the answer should be
$$1$$
Where did I fail?
 A: Alternatively you may observe that, for any differentiable function $f$, you have

$$
\frac{f(x)-f(a)}{x-a} \to f'(a).
$$ 

Then use it with $a=1$ and
$$
f(x)=\sqrt[3]{4-3x},\quad \quad f'(x)=-\frac{1}{(4-3 x)^{2/3}}
$$ giving
$$
\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)=-f'(1)=1.
$$
A: the problem is that you used 
(a-b)(a+b) to simplify a cubic root instead of a simple root
A: You mixed up square roots and cube roots:
$$(1-\sqrt[2]{a})*(1+\sqrt[2]{a})=1-a$$
$$(1-\sqrt[3]{a})*(1+\sqrt[3]{a})≠1-a$$
A: I think you overlooked this multiplication $(1+(4-3x)^{1/3})(1-(4-3x)^{1/3})$ which equals $1-(4-3x)^{2/3}$ not $1-(4-3x)$
A: $ (1-(4-3x)^(1/3))(1+(4-3x)^(1/3)) = 1 - (4-3x)^(2/3) $
I couldn't re-format it right but others beat me to it
A: If you let $t=4-3x,\;$ so $x=\frac{1}{3}(4-t)$,  you get
$\displaystyle\lim_{x\to 1}\frac{1-\sqrt[3]{4-3x}}{x-1}=\lim_{t\to1}\frac{1-\sqrt[3]{t}}{\frac{1}{3}(1-t)}=\lim_{t\to1}\frac{3(1-t^{1/3})}{(1-t^{1/3})(1+t^{1/3}+t^{2/3})}=\lim_{t\to1}\frac{3}{1+t^{1/3}+t^{2/3}}=1$
