How to solve parametrized limits?

Question

I am a bit confused at the moment. This exercise in particoular shattered my self confidence:
$$\lim\limits_{x\to 0}\frac{\sinh(x) + 1 - (1 + 3x)^{\frac13}}{\ln\left(1 - 2x^α\right) + 2x^3}$$ with $\alpha \in \mathbb R$
My questions are:
1)How do I solve this kind of limit? I tried with McLaurin (keeping the parameter) but it kind of "branches" into many sub-cases and I quckly got lost. Is there a general approach?
2) How can I check the solution? I normally use wolfram but it just says lim=0. The best I could come up with so far was geogebra, using the X of a poind as the parameter, but (according to geogebra) the limit changes if $\alpha$ is an integer or not (for $\alpha<-2$ at least), and t doesn't seem the ideal tool for this kind of situation

Answer 1

As I explained in my previous comment, I think the question is meaningful only for $\alpha>0$ and $x\to0^+$. As both numerator and denominator tend to $0$ the simplest way is to find the lowest order term when expanding both around $x=0$. You have $$\ln(1+t)=t-(1/2)t^2+\hbox{higher order terms},$$ so your denominator is $$ \ln(1-2x^\alpha)+2x^3=2x^3-2x^\alpha-2x^{2\alpha}+\hbox{higher order terms}. $$ The lowest order term in denominator is then $-2x^\alpha$ if $\alpha<3$, it is $2x^3$ if $\alpha>3$ and it is $-2x^6$ if $\alpha=3$, because in that case there is a cancellation. These are the three cases to consider. Expand now the numerator and draw your conclusions.