Is this allowed

[gloomy wedge]

Or should I evalute the whole limit for d/dh(f(x+h)-f(x)) and d/dh(h) as well???

[half knot]

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[dim smelt]

do you know that f is differentiable?

[gloomy wedge]

better way to see it

[dim smelt]

this works if f' is continuous

but you probably don't know that

[gloomy wedge]

fair enough, the question i got this from gave a function which was shown to be continuous

[dim smelt]

you can apply lHopital if you know the function in question is differentiable

but you are verifying differentiability by definition

so I suspect it is not given that f is differentiable

[gloomy wedge]

the question was to differentiate for f'(0) with differentiability by defintion, so it just seemed a bit contradictory to use l'Hopital without also using differentiability by definition for l'Hopitals as well

[dim smelt]

then no lhopital

calculate the limit for x=0

[gloomy wedge]

it was done earlier in the question to show that the function was continuous and then we had to differentiate by definition

[dim smelt]

yup

no lhopital

and be careful with this

this works for continuously differentiable functions

[gloomy wedge]

this was the function, which in step 1: show that f(x) is continuous and diffrerentialble in x = 0. step 2: use differentiability by definition to calculate f'(0)

[dim smelt]

if you show it's differentiable at 0, then it's also continuous

[gloomy wedge]

yeah, my question was more or less originally if I am asked to use differentiability by definition to solve f'(0) and then use l'Hopitals to calculate it. Would that imply that I use differentiability by definition be used for l'Hopitals?

[dim smelt]

no lhopital

you don't even know if you can use it

if you calculate this limit

then you already know what f'(0) is, if it exists

there's no need to recalculate it with lhopital

$$ \lim _{h\to 0} \frac{\frac{\sqrt{1+h} -\cos h}{h} - \frac{1}{2}}{h} $$

[wispy tusk]

aL

[gloomy wedge]

they kinda just applied l'Hopital to that limit twice

[dim smelt]

so verify the assumptions of lhopital theorem

before you apply it

[gloomy wedge]

so just that the function in the denominator and the numerator evaluates to the same indeterminate form?

[dim smelt]

i guess..whatever that means

are the limits of numerator and denominator both 0?

[gloomy wedge]

yes

[dim smelt]

are both numerator and denominator differentiable around 0?

[gloomy wedge]

yes

[dim smelt]

good, proceed

[gloomy wedge]

the question i have is pretty stupid ngl but its more if the semantics would imply this would be the way to go about applying l'Hopitals since you differentiate the numerator and denominator

if i am asked to solve f'(0) with differentiability by defintion

it would just become excessive lol

[dim smelt]

what the

where did all that come from?

[gloomy wedge]

ok tried to make it better to understand

[dim smelt]

how does it get so long tho

[gloomy wedge]

differentiability by defintion lol

very excessive

[dim smelt]

$$ \frac{d}{dh}2(\sqrt{1+h} -\cos h)-h = 2\left( \frac{1}{2\sqrt{1+h}} +\sin h \right)-1 $$

[wispy tusk]

aL

[dim smelt]

at this point you are expected to know that finite compositions of differentiable functions are again differentiable

[gloomy wedge]

the semantics of the problem could be interpreted as to only use differentiability by definition when differentiating which then would make using l'Hopitals very excessive

anyways this a very stupid question i gotta get studying done for a test tomorrow

[dim smelt]

[gloomy wedge]

thanks for your time to try to interpret this dumb question @dim smelt

[silk mist]

@gloomy wedge has given 1 rep to @dim smelt

[sage bison]

You don't have to be sorry if your question is trivial to others.

[gloomy wedge]

it's just dumb in a goofy way, rather than stupid

imo

[sage bison]

We are here to help at any matter (if it is in our capability) and as long as you really want to know something and aren't just trolling, we will be happy to help.

btw, please close the thread when you are done. You can do this with +close

[gloomy wedge]

+close