Find $\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$ Find the limit without using L'hopital 
$$\lim_{x \to 0 }\frac{(1+x)^{(1/2)} -1}{(1+x)^{(1/3)} -1}$$
I have tried multiply by the conjugate of the denominator and numerator but didn't work 
Any hint would be appreciated 
Thank you
 A: Let $1+x=(1+y)^6$.
The expression becomes
$\dfrac{(1+y)^3-1}{(1+y)^2-1}
=\dfrac{3y+3y^2+y^3}{2y+y^2}
=\dfrac{3+3y+y^2}{2+y}
\to \dfrac32
$.
Generalizations should be clear.
A: There are two conjugates you need to multiply:
\begin{align*}
\lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} &= \lim_{x\to0} \frac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} \cdot \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1} \\
&= \lim_{x\to0} \frac{(\sqrt{1 + x} - 1)(\sqrt{1 + x} + 1)}{(\sqrt[3]{1 + x} - 1)(\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1)} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\ 
&= \lim_{x\to0} \frac{\sqrt{1 + x}^2 - 1^2}{\sqrt[3]{1 + x}^3 - 1^3} \cdot \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\
&= \lim_{x\to0} \frac{\sqrt[3]{1 + x}^2 + \sqrt[3]{1 + x} + 1}{\sqrt{1 + x} + 1} \\
&= \frac{\sqrt[3]{1 + 0}^2 + \sqrt[3]{1 + 0} + 1}{\sqrt{1 + 0} + 1} = \frac{3}{2}.
\end{align*}
A: You can use the fact that for $x$ near $0$, 
$$(1+x)^\alpha = 1+\alpha x+o(x)$$
Then, in the limit $x\to0$, 
\begin{align}
\lim_{x\to0} \frac{(1+x)^{1/2}-1}{(1+x)^{1/3}-1} &= \lim_{x\to 0}\frac{1+x/2+o(x)-1} {1+x/3+o(x)-1} \\
&=\lim_{x\to0} \frac{x/2+o(x)}{x/3+o(x)} \\
&= \lim_{x\to 0} \frac{3}{2} \\
&=\frac{3}{2}
\end{align}
A: Using that $t^2-1=(t-1)(t+1)$ and $t^3-1=(t-1)(t^2+t+1)$ we have
$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}\cdot \dfrac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\cdot \dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}\right]$$ and thus
$$\lim_{x \to 0}\dfrac{\sqrt{x+1}-1}{\sqrt[3]{x+1} -1}=\lim_{x \to 0}\left[\dfrac{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}{\sqrt{x+1}+1}\right]=\dfrac 32.$$
A: Hint:
As lcm$(2,3)=6$
choose $\sqrt[6]{1+x}=y\implies\sqrt[2]{1+x}=y^3,\sqrt[3]{1+x}=y^2$
and as $x\to0,y\to1$
OR
choose $\sqrt[6]{1+x}=z+1\implies z\to0$
A: Let $y:=1+x;$  $y>0$.
$\dfrac{y^{1/2}-1}{y^{1/3}-1}=$
$\dfrac{y -1}{y^{1/2}+1}\dfrac{y^{2/3}+y^{1/3}+1}{y-1}.$
The limit $y \rightarrow 1$ is?
Used: 
$y-1=(y^{1/2}-1)(y^{1/2}+1)$, and
$y-1=(y^{1/3}-1)(y^{2/3}+y^{1/3} +1)$.
