Sandwich theorem limit problem

[issacnewton]

Homework Statement

Prove that
$$ \lim_{x\to 0} \sqrt{x^3+x^2}\; \sin\left(\frac{\pi}{x}\right) = 0 $$

using Sandwich theorem

Homework Equations

Sandwich Theorem

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

 

[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)

 

[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 ?

 

[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).

 

[issacnewton]

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

 

[Ray Vickson]

Homework Statement

Prove that
$$ \lim_{x\to 0} \sqrt{x^3+x^2}\; \sin\left(\frac{\pi}{x}\right) = 0 $$

using Sandwich theorem

Homework Equations

Sandwich Theorem

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

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

 

[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.

 

[Ray Vickson]

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.

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

 

[issacnewton]

Thanks Ray... makes sense