Calculating This Limit Question w/o L'Hospital

Question

$$\lim_{x\to 0} \frac{\sqrt[3]{x+5}-\sqrt[3]{5}}{x} = \frac{1}{3 \cdot 5^{2/3}}$$

I need to solve this without L'Hospital and I need something like $x^{1/3} = t^3$ or etc.

I need to use $a^3-b^3 = (a-b)(a^2+ab+b^2)$.

Thank you.

Answer 1

Solutions have been posted that (i) recognize the limit as a derivative and (ii) rationalize the numerator. We proceed, therefore, to present another way forward.

Here, we use the Generalized Binomial Theorem to write

$$\begin{align} (x+5)^{1/3}&=5^{1/3}\left(1+\frac x5\right)^{1/3}\\\\ &=5^{1/3}\left(1+\frac{x}{15}+O(x^2)\right)\tag 1 \end{align}$$

Using $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{(x+5)^{1/3}-5^{1/3}}{x}\right)=\lim_{x\to 0}5^{1/3}\left(\frac1{15}+O(x)\right)=\frac{1}{3\cdot 5^{2/3}}}$$

as expected!

Answer 2

As you already know which identity to use its pretty much trivial.

$\lim_{x\to 0}\dfrac{((x+5)^{1/3}-5^{1/3})((x+5)^{2/3}+5(x+5)^{1/3}+5^{2/3})}{x((x+5)^{2/3}+5(x+5)^{1/3}+5^{2/3})}=\dfrac{1}{5^{2/3}+5^{2/3}+5^{2/3}}$.

Answer 3

Why would anyone even try L'Hopital? The limit is just the definition of the derivative of $f(x) = (x+5)^{1/3}$ at $0.$