$$\lim_{x\rightarrow 0} \,\,\left( \sqrt[3]{1+2x+x^3} - \frac{2x}{2x+3} \right) ^ {\frac1{x^3}} $$
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?
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 $$\sqrt[3]{1.1}=1.032280\dots,\frac{3.2}{3.1}=1.032258\dots$$
Let $L$ be the limit, then taking the natural log, $$\begin{split}\log L&=\lim_{x\to0}\frac1{x^3}\log\left(\sqrt[3]{1+2x+x^3}-\frac{2x}{2x+3}\right)\\&\stackrel{\mathrm{L'H}}{=}\lim_{x\to0}\frac{1}{3x^2}\frac{1}{\sqrt[3]{1+2x+x^3}-(2x)/(2x+3)}\left(\frac{3x^2+2}{3(1+2x+x^3)^{2/3}}-\frac{6}{(2x+3)^2}\right).\end{split}$$ The latter limit is rather tedious to evaluate, but can be done. (If we should encounter indeterminate forms again, then just apply L'Hospital's rule again.) You should end up finding that $\log L=43/81$, after which exponentiating both sides gives the answer. The key idea here is taking the logarithm, which allows us to convert the given limit into the form required to apply L'Hospital's rule.
Taking the logarithm, what I should do is to consider $$\frac{\log \left(\sqrt[3]{x^3+2 x+1}-\frac{2 x}{2 x+3}\right)}{x^3}$$ and you need to apply L'Hospital rule three times (just because of the denominator). Not funny but doable. And the limit is effectively $\frac{41}{81}$.
Now, take the exponential of it.
