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

• Apr 7th 2010, 08:43 AM
frenchguy87
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.
• Apr 7th 2010, 11:44 AM
Failure
Quote:

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 $\displaystyle \lim_{x\to 0+}sin(1/x)$.
• Apr 7th 2010, 03:53 PM
frenchguy87
fixed!
• Apr 8th 2010, 08:15 AM
frenchguy87
does anyone know?
• Apr 8th 2010, 09:33 AM
Defunkt
Quote:

Originally Posted by frenchguy87
does anyone know?

Where are you having problems? This is rather straight forward.
• Apr 8th 2010, 01:54 PM
frenchguy87
Never mind I figured it out earlier.