<s>[INST] how to find the value of the $k$ to make $\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$ exist? the question described as follow:
$$\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=A$$
the $A$ is constant and $A\not=0$
and find the $k$ to make this limit exist.

and I did this:
$$let\space t\space be\space x^{1/15}\space, then \space x=t^{15}.$$
$$then \space \lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=\lim_{x\to 0} \frac{\sqrt[5]{t^{15}+t^{5}}-\sqrt[3]{t^{15}+t^{3}}}{t^{15k}}.$$
$$then \space \lim_{x\to 0} \frac{\sqrt[5]{t^{15}+t^{5}}-\sqrt[3]{t^{15}+t^{3}}}{t^{15k}}=\lim_{x\to 0} \frac{t\sqrt[5]{t^{10}+1}-t\sqrt[3]{t^{12}+1}}{t^{15k}}.$$
use taylor expantion:
$$then \space \lim_{x\to 0} \frac{t\sqrt[5]{t^{10}+1}-t\sqrt[3]{t^{12}+1}}{t^{15k}}=\lim_{x\to 0} \frac{t(1+\frac{1}{5}t^{10}+o(t^{10}))-t(1+\frac{1}{3}t^{12}+o(t^{12}))}{t^{15k}}.$$
$$then \space\lim_{x\to 0} \frac{t(1+\frac{1}{5}t^{10}+o(t^{10}))-t(1+\frac{1}{3}t^{12}+o(t^{12}))}{t^{15k}}=\lim_{x\to 0} \frac{\frac{1}{5}t^{11}+o(t^{11})}{t^{15k}}.$$
$$then \space\lim_{x\to 0} \frac{\frac{1}{5}t^{11}+o(t^{11})}{t^{15k}}=\lim_{x\to 0} \frac{\frac{1}{5}t^{11}}{t^{15k}}=\lim_{x\to 0} \frac{1}{5t^{15k-11}}=A.$$
since A is a non-zero constant, so the  $t^{15k-11}$ should be $t^{0}=1$.
then we get $15k-11=0$ and finally, we find $k=\frac{11}{15}$.

Am I right?
suppose I was right.
but unfortunately I found the image of $f(x)=\frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{\frac{11}{15}}}$ in quick graph app.
the value of $f(0)$ goes to $\infty$ instead any constant.
which is wrong, the app or me?
if I was wrong, how to find the right k?
 [/INST] It seems correct indeed
$$\sqrt[5]{x+\sqrt[3]{x}}\sim x^\frac1{15} \left(1+ \frac{x^\frac23}5 \right)=x^\frac1{15} + \frac{x^\frac{11}{15}}5 $$
$$\sqrt[3]{x+\sqrt[5]{x}}\sim x^\frac1{15} \left(1+\frac{x^\frac45}3\right)=x^\frac1{15} +\frac{x^\frac{13}{15}}3$$
thus
$$\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}\sim \frac{x^\frac{11}{15}}5 $$
and
$$\lim_{x\to 0} \frac{ \sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}} }{x^\frac{11}{15}}=\frac15$$
</s>