Limit of $\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot[{(x+\sin{\frac{1}{x}})}^{\frac{1}{3}} -x^{\frac{1}{3}}]}$ I need to find the limit of 
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
 A: Putting $\frac1x=h$
$$\lim_{x \to \infty}{x^{\frac{5}{3}}\cdot\left[{\left(x+\sin{\frac{1}{x}}\right)}^{\frac{1}{3}} -x^{\frac{1}{3}}\right]}$$
$$=\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}$$
Method $1:$
Using Binomial Expansion (1, 2) , $(1+x)^n=1+nx+O(x^2)$ and  as $\lim_{h\to0}\frac{\sin h}h=1\implies \sin h $ is $O(h)$
$$\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}=\lim_{h\to0}\frac{\frac13h\sin h+O(h^4)}{h^2}=\frac13\lim_{h\to0}\frac{\sin h}h$$
Method $2:$
As $a^3-b^3=(a-b)(a^2-ab+b^2),$
$$\lim_{h\to0}\frac{(1+h\sin h)^{\frac13}-1}{h^{\left(\frac53+\frac13\right)}}$$
$$=\lim_{h\to0}\frac{(1+h\sin h)-1}{h^2\{(1+h\sin h)^{\frac23}+(1+h\sin h)^{\frac13}+1\}}$$
$$=\lim_{h\to0}\frac{\sin h}h\cdot\frac1{\lim_{h\to0}\{(1+h\sin h)^{\frac23}+(1+h\sin h)^{\frac13}+1\}}\text{ as }h\ne0 \text{ as }h\to0$$
$$=1\cdot\frac1{\{(1+0)^{\frac23}+(1+0)^{\frac13}+1\}}$$
A: Elementary hint:
Multiply the function by $$1=\frac{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}{\sqrt[3]{(x+\sin(1/x))^2}+\sqrt[3]{x(x+\sin(1/x))}+\sqrt[3]{x^2}}$$ and use the basic well known fact that $a^3-b^3=(a-b)(a^2+ab+b^2)$ to get $1/3$.
A: Hint: Write $$x+\sin \frac1x = x\left(1+\frac1x \sin \frac1x\right)$$ and use
$$\sin \frac1x=\frac1x - \frac{1}{3!x^3} + \frac{1}{5!x^5}+ \ldots, \quad x \to \infty.$$
A: $$Limit=\lim_{x\rightarrow\infty}x^2[(1+\frac{\sin \frac{1}{x}}{x})^\frac{1}{3}-1]=\lim_{x\rightarrow\infty}x^2\frac{\sin \frac{1}{x}}{3x}=\frac{1}{3}$$
