<s>[INST] How to find the limit of this function using L'Hospital's rule? 
$$\lim_{x\rightarrow 0} \,\,\left( \sqrt[3]{1+2x+x^3} - \frac{2x}{2x+3}   \right) ^ {\frac1{x^3}} $$

I have already tried several options, but the only answer I have gotten so far is
$e^{\infty}$, which is incorrect. The correct answer is $e^\frac{43}{81}$, which is easy to get by using Taylor series, but our task was to get the same one by using L'Hospital's rule. Can you help me with it?
 [/INST] The standard approach here is to take logarithm and then evaluate the limit. If $f(x) $ denotes the expression under limit and it tends to a limit $L$ then
\begin{align}
\log L&=\lim_{x\to 0}\frac{1}{x^3}\log\left(\sqrt[3]{1+2x+x^3}-\frac{2x}{2x+3}\right)\notag\\
&=\lim_{x\to 0}\frac{g(x)}{x^3}\cdot\frac{\log(1+g(x))}{g(x)}\notag\\
&=\lim_{x\to 0} \frac{g(x)} {x^3}\notag\\
&=\lim_{x\to 0}\frac{1}{x^3}\left(\sqrt[3]{1+2x+x^3}-1-\frac{2x}{2x+3}\right)\notag
\end{align}
The algebraic limit at the end can be evaluated using standard techniques. Here is one approach which requires some algebra. If $$A=\sqrt[3]{1+2x+x^3},B=\frac{4x+3}{2x+3}$$ then $A,B$ both tend to $1$ and $$A^3-B^3=x^3+\frac{(1+2x)(2x+3)^3-(4x+3)^3}{(2x+3)^3}=x^3+\frac{16x^3(1+x)} {(2x+3)^3}$$ so that $(A^3-B^3)/x^3\to 43/27$. And therefore $$\frac{A-B} {x^3}=\frac{A^3-B^3}{x^3}\cdot\frac{1}{A^2+AB+B^2}\to \frac{43}{81}$$ Thus the desired limit is $L=e^{43/81}$.
Advanced tools like L'Hospital's Rule and Taylor series are often unnecessary for simple limit problems and instead a little bit of algebra helps.

Incidentally this exercise helps us to know that $(3+2x)/(3+x)$ is a very good approximation to $\sqrt[3]{1+x}$ for small values of $x$. For example if $x=0.1$  then we have $$\sqrt[3]{1.1}=1.032280\dots,\frac{3.2}{3.1}=1.032258\dots$$
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