Ask your own question, for FREE!
Mathematics 27 Online
OpenStudy (anonymous):

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?

OpenStudy (zarkon):

f''(x) = 10 is not true in general

OpenStudy (anonymous):

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?

OpenStudy (zarkon):

\[f(x)=5x^2e^x\] 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 $$\frac{f''(x)}{2}=5$$

OpenStudy (zarkon):

you don't need to apply L'Hospitals rule two times

OpenStudy (anonymous):

How do I know the value of f'(0) then? It must be a very simple thing that I didn't get.

OpenStudy (zarkon):

do we know if f'(x) is continuous at zero?

OpenStudy (anonymous):

f'(x) I don't know, but: \[f : R \rightarrow R\]

OpenStudy (zarkon):

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 $$ \lim_{x \rightarrow 0} f'(x)=0$$

OpenStudy (anonymous):

But then the limit still equals 5?

OpenStudy (zarkon):

yes

OpenStudy (zarkon):

for the original problem

OpenStudy (anonymous):

Can I say f'(x) = 10x because of this? \[\lim_{x \rightarrow 0} \frac{f'(x)}{2x} = 5\]

OpenStudy (zarkon):

no

OpenStudy (zarkon):

you know nothing else about f'(x) except what happens near zero

OpenStudy (anonymous):

But I don't know what happens near zero. :p I'm trying to understand what you said about the limit converging.

OpenStudy (zarkon):

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

OpenStudy (zarkon):

therefore we must have \[\lim_{x\to 0}f'(x)=0\]

OpenStudy (zarkon):

and if f'(x) was continuous then we get \[0=\lim_{x\to 0}f'(x)=f'(\lim_{x\to 0}x)=f'(0)\]

OpenStudy (anonymous):

Oh, I think I got it. (not this last part)

OpenStudy (zarkon):

the last thing I wrote is confusing? $$0=\lim_{x\to 0}f'(x)=f'(\lim_{x\to 0}x)=f'(0)$$

OpenStudy (anonymous):

Yes, never seen that before.

OpenStudy (zarkon):

limits pass through continuous functions

OpenStudy (zarkon):

if F is a continuous function at a then $$\lim_{x\to a}F(x)=F(a)$$

OpenStudy (anonymous):

Yeah, alright.

OpenStudy (zarkon):

so if f'(x) is continuous at 0 then $$\lim_{x\to 0}f'(x)=f'(0)$$

OpenStudy (zarkon):

all good then?

OpenStudy (anonymous):

Yeah, yeah. Very obvious. :p

OpenStudy (zarkon):

cool...time to spend some time with the Wife ...later

OpenStudy (anonymous):

Heh... Thank you very much, Zarkon!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
euphoriiic: I drew more people !! ud83eudd17.
2 minutes ago 41 Replies 3 Medals
euphoriiic: I drew markk ud83eudd17.
4 hours ago 34 Replies 2 Medals
LOLIAteYourMom: u2212243 = u22129(10 + x)
7 hours ago 0 Replies 0 Medals
Bubblezz: Art for @jeromeccv
7 hours ago 13 Replies 3 Medals
Bubblezz: Art for @jeromecv
11 hours ago 0 Replies 0 Medals
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!