# Thread: Limit Question

1. ## Limit Question

$\lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}$

2. Hello, treetheta!

Edit: Made a blunder . . . will correct it now.

We use: . $(a - b)(a^2+ab+b^2) \:=\:a^3-b^3$

$\lim_{x \to 0} \frac{(2x+1)^{\frac{1}{3}}-1}{x}$

Multiply top and bottom by: $(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1$

. . $\frac{(2x+1)^{\frac{1}{3}} - 1}{x}\cdot\frac{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}$ . $= \;\frac{(2x+1) - 1}{x\left[(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1\right]}$

. . $= \;\frac{2{\color{red}\rlap{/}}x}{{\color{red}\rlap{/}}x\left[(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1\right]} \;=\;\frac{2}{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}$

We have: . $\lim_{x\to0}\left[\frac{2}{(2x+1)^{\frac{2}{3}} + (2x+1)^{\frac{1}{3}} + 1}\right] \;=\;\frac{2}{1^{\frac{2}{3}} + 1^{\frac{1}{3}} + 1} \;=\;\frac{2}{3}$

3. how did this

$\lim_{x \to \o} \frac{(2x+1)^{1/3}-1}{x}$

go to

$\lim_{x \to \o} \frac{(2x+1)^{1/3}-1}{1}$

what happend to the x in the denominator

ya, if u add the x, the answer becomes 2/3 thankssss ^^

4. Is the limit supposed to be going to infinity or zero?

5. It does, it's going to zero

i put o instead of 0 sorry

6. Originally Posted by treetheta
$\lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}$
l'Hopital's Rule shows that the limit is equal to 2/3.

Another way to do it is to substitute $t = (2x + 1)^{1/3} \Rightarrow x = \frac{t^3 - 1}{2}$. Then the limit becomes:

$\lim_{t \rightarrow 1} \frac{2(t-1)}{t^3 - 1}$.

Factorise the denominator, cancel the common factor of (t-1) in the numerator and denominator and then take the limit.

7. Originally Posted by treetheta
$\lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x}$
Yet another way : Let $\begin{array}{rl}f:\mathbb{R} &\mapsto \mathbb{R}\\ x & \mapsto \sqrt[3]{2x+1}\end{array}$. $f$ is differentiable on $\mathbb{R}-\left\{-\tfrac12\right\}$ and $f'(x)=\frac{2/3}{(2x+1)^{2/3}}$.

$\lim_{x \to 0} \frac{\sqrt[3]{2x+1}-1}{x}=\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}$

That's $f'(0)$...