I need to find the values of $a$ for which the following limit is $0$ :
$$L = \lim _{x\rightarrow 0^{+}}\frac{\sin(\log( 1+3x)) -e^{3x} +\cos x}{(\sin x)^{3\alpha }}$$
It is an indeterminate form of the type $\frac{0}{0}$.
I think I need to use l'Hopital's rule, but those calculations become cumbersome & difficult and I can't figure out anything. Is there perhaps something I'm missing? I know the limit $L$ is $0$ when the numerator is $0$, but I can't figure out how to find $a$ from that.
Thanks to anyone who decides to help me :).
We must have $x \rightarrow 0^{+}$ because negative values will give Imaginary or Complex Denominator in general.
By "Inspection" :
We can see that when $a$ is negative , we can use the reciprocal & move that term to Numerator , hence it will be $0/1$ , hence "$a$ negative & limit $L=0$" is a Solution.
We can then see that $a=0$ will still make it $0/1$ , hence "$a=0$ & limit $L=0$" is a Solution.
There might be some Positive values where Numerator is still $0$ when the largest Common $x$ Power is cancelled between Numerator & Denominator. Beyond that largest Common $x$ Power , Denomintor will be $0$. To get that Particular Power , we will have to use Series Expansion.
Series Expansion :
Now the Positive $a$ Case is $0/0$. We can use Series Expansion to get that the Numerator is $-19x^2/2+\text{higher-terms}$ , while Denominator is $x^{3a}+\text{higher-terms}$.
Over-all , it will be $-19x^{2-3a}/2+\text{higher-terms}$.
Existence of the limit indicates that $2-3a=0$ to give Constant Initial term , where the higher terms will be $0$ at $x=0$.
Hence $a=2/3$ will work & limit will be $L=-19/2$.
The last Solution is then "$a=2/3$ & limit $L=-19/2$" covering all Cases.
All smaller $a<2/3$ values will give $0$ limit.
All larger $a>2/3$ values will give $\infty$ limit.