I have a problem that asks the value of f'(0), and states f(0) = 0 and \[\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = 5\] This is what I have done: 1. As the limit is a zero/zero situation, I applied L'Hôpital's rule, so: \[\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = \lim_{x \rightarrow 0} \frac{f'(x)}{2x}\] 2. As the denominator still "equals" to zero, I applied L'Hôpital's rule again, so: \[\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = \lim_{x \rightarrow 0} \frac{f''(x)}{2} = 5\] 3. Then, f''(x) = 10. Integrating, I get f'(x) = 10x and from there I calculate: \[f'(0) = 10(0) = 0\] Is this right?
f''(x) = 10 is not true in general
Why not? Because I do not know what f'(x) is? Not knowing what f'(x) really is (but knowing the limit equals 5), can I apply L'Hôpital's rule the second time?
\[f(x)=5x^2e^x\] satisfies the conditions of your problem and $$f''(x)\neq10$$ you just have that in the limit you get 10 $$\lim_{x \rightarrow 0} \frac{f''(x)}{2} = 5$$ you don't have $$\frac{f''(x)}{2}=5$$
you don't need to apply L'Hospitals rule two times
How do I know the value of f'(0) then? It must be a very simple thing that I didn't get.
do we know if f'(x) is continuous at zero?
f'(x) I don't know, but: \[f : R \rightarrow R\]
if f'(x) is continuous then f'(0)=0 since the only way for this limit \[\lim_{x \rightarrow 0} \frac{f'(x)}{2x}\] to converge is if $$ \lim_{x \rightarrow 0} f'(x)=0$$
But then the limit still equals 5?
yes
for the original problem
Can I say f'(x) = 10x because of this? \[\lim_{x \rightarrow 0} \frac{f'(x)}{2x} = 5\]
no
you know nothing else about f'(x) except what happens near zero
But I don't know what happens near zero. :p I'm trying to understand what you said about the limit converging.
if \[\lim_{x\to 0}f'(x)\neq0\] then \[\lim_{x\to 0}\frac{f'(x)}{2x}\] would not exist since the bottom would be going to zero but the top would not
therefore we must have \[\lim_{x\to 0}f'(x)=0\]
and if f'(x) was continuous then we get \[0=\lim_{x\to 0}f'(x)=f'(\lim_{x\to 0}x)=f'(0)\]
Oh, I think I got it. (not this last part)
the last thing I wrote is confusing? $$0=\lim_{x\to 0}f'(x)=f'(\lim_{x\to 0}x)=f'(0)$$
Yes, never seen that before.
limits pass through continuous functions
if F is a continuous function at a then $$\lim_{x\to a}F(x)=F(a)$$
Yeah, alright.
so if f'(x) is continuous at 0 then $$\lim_{x\to 0}f'(x)=f'(0)$$
all good then?
Yeah, yeah. Very obvious. :p
cool...time to spend some time with the Wife ...later
Heh... Thank you very much, Zarkon!
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