<s>[INST] 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
 [/INST] 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*}
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