Solve the limit $\lim_{x \to 3}\frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}$ with no l'Hopital rule

Question

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 $x=4$ $$\frac{1}{3\cdot4^{\frac{2}{3}}} = \frac{1}{3\cdot2^{\frac{4}{3}}} = \frac{1}{6\sqrt[3]{2}}$$

Answer 1

Hint: This limit is simply the derivative of something. Can you figure it out from there?

(If it helps, make the substitution $x = 3+h$ so that we have a limit as $h \to 0$).

Answer 2

Hint: The fraction is of the form $$\frac{a-b}{a^3-b^3}$$ so you can factor the denominator and cancel the factor that goes to zero.

Can you find the appropriate choices for $a$ and $b$?

Answer 3

Use the propertie: $$a^3-b^3=(a-b)(a^2+ab+b^2). $$

Indeed: $$\lim_{x\rightarrow 3} \frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}=\lim_{x\rightarrow 3} \frac{\sqrt[3]{x+1}-\sqrt[3]{4}}{x-3}\frac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}\sqrt[3]{4}+\sqrt[3]{4^2}}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}\sqrt[3]{4}+\sqrt[3]{4^2}}=$$ $$=\lim_{x\rightarrow 3}\frac{(\sqrt[3]{x+1})^3-(\sqrt[3]{4})^2}{(x-3)(\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}\sqrt[3]{4}+\sqrt[3]{4^2})} =$$ $$=\lim_{x\rightarrow 3}\frac{1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}\sqrt[3]{4}+\sqrt[3]{4^2}}=\frac{1}{\sqrt[3]{4^2}+\sqrt[3]{4}\cdot\sqrt[3]{4}+\sqrt[3]{4^2}} =\frac{1}{3\sqrt[3]{16}}.$$

Answer 4

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 $$\lim_{y \to a}\frac{y^{n} - a^{n}}{y - a} = na^{n - 1}$$