Calculate limit of piecewise function

[lep11]

Homework Statement

a.) Let $f,g:ℝ→ℝ$ such that $g(x)=sin x$ and $f(x)= \left\{ \begin{array}{ll} x^2, x∈ℚ \\ 0 , x∈ℝ\setminusℚ \\ \end{array} \right.$. Calculate $\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$.

b.) Why l'Hospital rule cannot be applied here?

The Attempt at a Solution

$\lim_{x \rightarrow 0} sinx=0$[/B]

$\lim_{x \rightarrow 0} f(x) =\left\{\begin{array}{ll} 0, x∈ℚ \\ 0 , x∈ℝ\setminusℚ \\ \end{array} \right.$

How to begin? I suppose I cannot apply l'Hospital.

 

[andrewkirk]

If you think about the values of $f(x)$ just on rational values of $x$, it should become apparent what the limit must be, if there is a limit. Call that number $l$. Then prove that $l$ actually is the limit by an explicit $\epsilon$-$\delta$ argument.

 

[lep11]

If you think about the values of $f(x)$ just on rational values of $x$, it should become apparent what the limit must be, if there is a limit. Call that number $l$. Then prove that $l$ actually is the limit by an explicit $\epsilon$-$\delta$ argument.

The limit of $\frac{f(x)}{g(x)}$ is 0, right?

 

[Ray Vickson]

Homework Statement

a.) Let $f,g:ℝ→ℝ$ such that $g(x)=sin x$ and $f(x)= \left\{ \begin{array}{ll} x^2, x∈ℚ \\ 0 , x∈ℝ\setminusℚ \\ \end{array} \right.$. Calculate $\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}$.

b.) Why l'Hospital rule cannot be applied here?

The Attempt at a Solution

$\lim_{x \rightarrow 0} sinx=0$[/B]

$\lim_{x \rightarrow 0} f(x) =\left\{\begin{array}{ll} 0, x∈ℚ \\ 0 , x∈ℝ\setminusℚ \\ \end{array} \right.$

How to begin? I suppose I cannot apply l'Hospital.

l'Hospital's rule does not apply because $f(x)$ is not a differentiable function.

 

[lep11]

l'Hospital's rule does not apply because $f(x)$ is not a differentiable function.

Yes, $f$ is continuous and differentiable only at x=0 which is not sufficient.

 

[Ray Vickson]

$f$ is continuous and differentiable only at x=0.

Yes, $f$ is continuous and differentiable only at x=0 which is not sufficient.

Right, because l'Hospital's rule talks about $\lim_{x \to 0} f'(x)$, not just $f'(0)$ (at least as it is usually presented).

 

[andrewkirk]

The limit of $\frac{f(x)}{g(x)}$ is 0, right?

You are correct. So now you need to prove it, using an $\epsilon$-$\delta$ argument. That is, prove that, for any $\epsilon>0$ you can find a $\delta>0$ such that $|x|<\delta\Rightarrow\frac{f(x)}{g(x)}<\epsilon$.

 

[lep11]

If $x∈ℝ\setminusℚ$, then $\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x \rightarrow 0}\frac{0}{sinx}=\lim_{x \rightarrow 0}0=0$ That's trivially true.

If $x∈ℚ$, then $\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x \rightarrow 0}\frac{x^2}{sinx}=0$ (an educated guess)

Proof: Let $ε>0$ arbitrarily. Let $δ=+\sqrt(ε|sinx|)$

$|\frac{x^2}{sinx}|=\frac{x^2}{|sinx|}<ε$, when $|x|<δ=\sqrt(ε|sinx|)$.

My proof doesn't look very elegant, does it?

 

[andrewkirk]

Proof: Let $ε>0$ arbitrarily. Let $δ=+\sqrt(ε|sinx|)$

$\delta$ cannot be a function of $x$, as $x$ is not defined in that context. You either need a more careful $\epsilon$-$\delta$ argument, or you can use l'Hopital on $\frac{x^2}{\sin x}$, use that to infer an $\epsilon$-$\delta$ relationship and apply that to the restricted domain of $\mathbb{Q}$.