<s>[INST] how I can simplify this limit? So I got $f(x) = 1+\sin^2(x)-e^{x^{2}}$
and the limit :
$$\lim_{x\to 0^+} \cfrac{\ln(1+6x)f(x)}{x^a\cos(f(x))}$$
I have to simplify this limit :(in a form like  $\lim_{x\to o} x^{n-a}$ , or something similar)


*

*I noticed that $\cos(f(x))$    for x->$0$ = $1$

*I can see => $\ln(1+6x)$  like   $\cfrac{6x\ln(1+6x)} {6x}$ that for $x→0$ is equal to $1$ (and I remain with $6x$)


now I have :
$6\lim_{x\to 0^+} {x^{1-a}f(x)}$
Can you help me simplify $f(x)$ that remains in the limit.
 [/INST] By using Taylor expansions, we have that as $x\to 0^+$,
$$f(x) = 1+\sin^2(x)-e^{x^{2}}=1+(x-x^3/6+O(x^5))^2-(1+x^2+x^4/2+O(x^6))\\=
1+x^2-x^4/3-1-x^2-x^4/2+O(x^5)=-5x^4/6+O(x^5).$$
Hence
$$\frac{\ln(1+6x)\cos(f(x)}{x^af(x)}=\frac{(6x+O(x^2))\cos(-5x^4/6+O(x^5))}{x^a(-5x^4/6+O(x^5))}\\
=\frac{(6x+O(x^2))\cdot(1+O(x^8))}{-5x^{4+a}/6+O(x^{5+a})}
\sim-\frac{36}{5x^{3+a}}.$$
</s>