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?

Question

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.

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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 $k=\frac{11}{15}$.

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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?

Answer 1

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

$$\lim_{x\to 0} \frac{ \sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}} }{x^\frac{11}{15}}=\frac15$$

Answer 2

Let $x=t^{15}$. Then

$$\lim_{x\to 0} \frac{\sqrt[5]{x+\sqrt[3]{x}}-\sqrt[3]{x+\sqrt[5]{x}}}{x^{k}}=\lim_{t\to 0} \frac{\sqrt[5]{t^{15}+t^5}-\sqrt[3]{t^{15}+t^3}}{t^{15k}}=\lim_{t\to 0} \frac{\sqrt[5]{t^{10}+1}-\sqrt[3]{t^{12}+1}}{t^{15k-1}}$$

By Taylor the first terms of the development will be $t^{10}$ and $t^{12}$ (to a nonzero factor) and you need

$$15k-1=10.$$