# How to solve parametrized limits?

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

• Not sure, what your confusion is. You should treat $\alpha$ as a constant for the sake of calculating the limit. In fact, the calculated limit would be either a constant number (assuming the limit exists) or a function of $\alpha$. An analogous in science is what we call "sensitivity analysis". You transform a set of inputs into an output using a transformation method and then test the variations in output if one or more of the input variables are tweaked (hence the name sensitivity analysis) – Deepak Gupta Sep 9 '15 at 9:05
• $\lim_{x\to0}1-2x^\alpha$ is $-\infty$ if $\alpha<0$ and $-1$ if $\alpha=0$. Since this is the argument of a logarithm, you must then consider only the case $\alpha>0$. – Aretino Sep 9 '15 at 9:13
• I know it needs to be treated as a constant, but for example studying the dervative of $\ln(1-2x^\alpha)$ is very different for $\alpha=1$ or $\alpha \neq 1$ – Riccardo Vailati Sep 9 '15 at 9:13
• Also, for odd negative values of $\alpha$ the limit exists for $x→0^-$ – Riccardo Vailati Sep 9 '15 at 9:19
• For solution checking, you can still use wolfram, but using Assumptions to verify each branch of your solution. – Terence Hang Sep 9 '15 at 9:44

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.