Solve the limit $\lim_{x \to 3}\frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}$ with no l'Hopital rule Solve the limit without using de l'Hopital rule: $$\lim_{x \to 3}\frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}$$
It looks like classical example of squeeze theorem, but I completely can't develop the intuition how to estimate lower and upper bounds. Any help would be appreciated.
Edit (solution):
According to hint of user @Omnomnomnom, we notice that the limit is derivative.
$$x=h+3$$
$$\lim_{h \to 0} \frac{\sqrt[3]{4+h}-\sqrt[3]{4}}{h}$$
It is derivative of $\sqrt[3]{x}$ in the point $x=4$. Thus:
$$\frac{d}{dx}x^{\frac{1}{3}} = \frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}}$$
Substituting the point $x=4$
$$\frac{1}{3\cdot4^{\frac{2}{3}}} = \frac{1}{3\cdot2^{\frac{4}{3}}} = \frac{1}{6\sqrt[3]{2}}$$

Let us put $y = x + 1$ so that $y \to 4$ and then we have $$\begin{aligned}L &= \lim_{x \to 3}\frac{\sqrt[3]{x + 1} - \sqrt[3]{4}}{x - 3}\\
&= \lim_{y \to 4}\frac{y^{1/3} - 4^{1/3}}{y - 4}\\
&= \frac{1}{3}\cdot 4^{(1/3) - 1} = \frac{1}{3\cdot 4^{2/3}} = \frac{1}{3\sqrt[3]{16}}\end{aligned}$$ We have used the formula $$\lim_{y \to a}\frac{y^{n} - a^{n}}{y - a} = na^{n - 1}$$
