How to evaluate the limit $\lim_{x\to 0} ((1+2x)^{1/3 }-1)/ x$ without using the l'Hospital's rule?

Question

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

Please do not use the l'hospital's rule as I am trying to solve this limit without using that rule... to no avail...

Answer 1

Hints:

As $x\to 0$, $\sqrt[3]{1+2x}-1\sim \frac23x$

In fact: $\sqrt[n]{1+x}-1\sim \frac xn$ as $x\to 0$.

Answer 2

In my old high school days, they used to call this ''multiply by the conjugate'' (this may make more sense later on). $$ \frac{(1+2x)^{1/3} - 1}x = \frac{[(1+2x)^{1/3} - 1][(1+2x)^{2/3} + (1+2x)^{1/3} + 1]}{x[(1+2x)^{2/3} + (1+2x)^{1/3} + 1]} \\ = \frac{2}{(1+2x)^{2/3} + (1+2x)^{1/3} + 1} \\ \underset{x \to 0}{\longrightarrow} \frac 23. $$ Added : For those who are learning Galois theory, the polynomial $y^3-a = (y-a)(y^2+ya+a^2)$ factors like this, and the root $y$ has Galois conjugates $\rho y$ and $\rho^2 y$, where $\rho$ is a third root of unity. We are sort of multiplying by these two conjugates, after correct interpretation.

Hope that helps,

Answer 3

Hint: Put $2x+1 = y^{3}$. And note that as $x \to 0 \implies y \to 1$. Then the limit becomes

\begin{align*} \lim_{x \to 0} \frac{(2x+1)^{1/3}-1}{x} &= \lim_{y \to 1} \frac{y-1}{\frac{y^{3}-1}{2}}\\ &= \lim_{y \to 1} \frac{2}{y^{3}-1} \times (y-1)\\ &= \lim_{y \to 1} \frac{2}{y^{2}+y+1} = \frac{2}{3} \end{align*}

Answer 4

I assume you're trying to avoid using calculus at all, as though you're trying to differentiate a function? Then let's try something completely different.

Let $u={\left(1+2x\right)}^{1/3}$. So $x=\frac{1}{2}\left(u^3-1\right)$, and $u \rightarrow 1$ as $x \rightarrow 0$. Then we have: $$\lim_{x \rightarrow 0}{\frac{\left(1+2x\right)^{1/3}-1}{x}}=\lim_{u \rightarrow 1}{\frac{u-1}{\left(\frac{u^3-1}{2}\right)}}=2\lim_{u \rightarrow 1}{\frac{u-1}{u^3-1}}$$ And, niftily enough, you might know that $u^3-1=\left(u-1\right)\left(u^2+u+1\right)$ and hopefully you can see how useful that is.

Answer 5

Just in the same spirit as Paul, remember that, for small values of $y$, $$(1+y)^n=1+n y+\frac{1}{2} n(n-1) y^2+\frac{1}{6}n (n-1) (n-2) y^3+O\left(y^4\right)$$ So, replace $n$ by $\frac{1}{3}$ and $y$ by $2x$ and obtain $$(1+2x)^{\frac{1}{3}}=1+\frac{2 x}{3}-\frac{4 x^2}{9}+\frac{40 x^3}{81}+O\left(x^4\right)$$ $$\frac{(1+2x)^{\frac{1}{3}}-1}{x}=\frac{2}{3}-\frac{4 x}{9}+\frac{40 x^2}{81}+O\left(x^3\right)$$ which shows the limit as well as the manner it is approached

Answer 6

Since, $x\to 0 \space => |2x|<1$ hence, we can binomial expansion of $(1+2x)^{\frac{1}{3}}$ $$\lim_{x\to 0}\frac{(1+2x)^{\frac{1}{3}}-1}{x}$$ $$=\lim_{x\to 0}\frac{\left(1+\left(\frac{1}{3}\right)\frac{(2x)}{1!}+\left(\frac{1}{3}\right)\left(\frac{1}{3}-1\right)\frac{(2x)^2}{2!}+..........\infty \right)-1}{x}$$ $$=\lim_{x\to 0}\left(\left(\frac{1}{3}\right)\frac{(2)}{1!}+\left(\frac{1}{3}\right)\left(\frac{1}{3}-1\right)\frac{(4x)}{2!}+..........\infty \right)=\left(\frac{2}{3}+0\right)=\frac{2}{3}$$

Answer 7

I have an easier way. Let $f(x) = (1+2x)^{\frac{1}{3}}$. $f(0) = 1$ So this is $f'(x)|_{x=0}$. $f'(x) = \frac{2}{3} \frac{1}{(1+2x)^{\frac{2}{3}}}$. $f'(0) = \frac{2}{3}$. Therefore the answer is $\frac{2}{3}$.