# Thread: L'Hospital's Rule to show non-existence of limits

1. ## L'Hospital's Rule to show non-existence of limits

Let $\displaystyle f(x)=x^2\sin{1/x}$ for $\displaystyle 0<x\leq 1$ and $\displaystyle f(0)=0$ , and let $\displaystyle g(x)=x^2$ for $\displaystyle x\in [0,1]$ .
Then both f and g are differentiable on [0,1] and $\displaystyle g(x)>0$ for $\displaystyle x\neq 0$ .
Show that $\displaystyle \lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x)$ and that $\displaystyle \lim_{x \to 0} f(x)/g(x)$ does not exist.

2. Originally Posted by frenchguy87
Let $\displaystyle f(x)=x^2\sin{1/x}$ for $\displaystyle 0<x\leq 1$ and $\displaystyle f(0)=0$ , and let $\displaystyle g(x)=x^2$ for $\displaystyle x\in [0,1]$ .
Then both f and g are differentiable on [0,1] and $\displaystyle g(x)>0$ for $\displaystyle x\neq 0$ .
Show that $\displaystyle \lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x)$ and that $\displaystyle \lim_{x \to 0} f(x)/g(x)$ does [?] exist.
But surely, there must be a typo in the statement of your problem: $\displaystyle \lim_{x\to 0+}f(x)/g(x)$ does not exist, since it boils down to $\displaystyle \lim_{x\to 0+}sin(1/x)$.

3. fixed!

4. does anyone know?

5. Originally Posted by frenchguy87
does anyone know?
Where are you having problems? This is rather straight forward.

6. Never mind I figured it out earlier.