# Sandwich theorem limit problem

1. Dec 27, 2016

### issacnewton

1. The problem statement, all variables and given/known data
Prove that
$$\lim_{x\to 0} \sqrt{x^3+x^2}\; \sin\left(\frac{\pi}{x}\right) = 0$$

using Sandwich theorem
2. Relevant equations
Sandwich Theorem

3. The attempt at a solution
Now we know that sine function takes values between -1 and 1. $-1 \leqslant \sin\left(\frac{\pi}{x}\right) \leqslant 1$. So we can multiply this by $\sqrt{x^3+x^2}$ on both sides. But I want to show that $\sqrt{x^3+x^2} \geqslant 0$ near $x = 0$. So I am stuck at this point. Any guidance will help.

Thanks

2. Dec 28, 2016

### Delta²

For x>0 its easy. Now for x<0 if you take for example $-1<x<0$ to prove that $x^3>-x^2$. Hint: try multiplying $-1<x<0$ by the proper positive term)

3. Dec 28, 2016

### issacnewton

Though I am solving this problem from Stewart's Calculus book, I am trying to be rigorous here. In the statement of the Sandwich theorem, we define an interval $I$ having $0$ as its limit point. Now if we take $I = [-1,1]$, then $0$ is definitely the limit point of $[-1,1]$. Now we need to show that for every x in $I$ not equal to $0$, we have
$$-\sqrt{x^3 + x^2} \leqslant \sin\left(\frac{\pi}{x}\right) \leqslant \sqrt{x^3 + x^2}$$

For this, we will first need to show that $\sqrt{x^3 + x^2} \geqslant 0$ for all $x$ in $[-1,1]$. Am I on right track ?

4. Dec 28, 2016

### Delta²

yes you are on the right track ( you just forgot to multiply by the $\sqrt{x^3+x^2}$ term in the middle of the double inequality).

5. Dec 28, 2016

### issacnewton

Oh I missed that.. Ok I will continue working on this.... thanx

6. Dec 28, 2016

### Ray Vickson

There is nothing to prove; that is the very definition of the function $\sqrt{...} .$

7. Dec 28, 2016

### issacnewton

Ray, we need to be sure that $x^3+x^2 \geqslant 0$. If this is not true, then there might be problem. Thats why I am tried to come up with an interval over which, we have $x^3+x^2 \geqslant 0$, and this interval also has $0$ as its limit point.

8. Dec 28, 2016

### Ray Vickson

For $0 \leq |x| \leq 1$ we have $|x|^3 \leq |x|^2 = x^2$.

9. Dec 28, 2016

### issacnewton

Thanks Ray... makes sense