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derivable or differentiable? careful with the jargon :P
derivable ! copied from my buk ! :p
i think both means the same
Thats ok ! Solve d prob ! :p
it definitely does not mean the same. if that's a calculus book it's a mistake.
malcom---its derivable ! No prob wd d jargon -solve it plzzz !
Woh Hansane wali baat thi anjali
use definition of continuity
I dint gt u ? Ur nt indian rite ? And plz no hints fr ques - straightforward explanatory ans ! I would be really obliged ! :p
anjali, mai apne khaane kaa anand le rahaa hoon.
anjali, one minute please
somewhere i had read .. the product of two continuous function is a continuous function is a continuous function ... somehow i can't find it
Hmmm, ok , search , then let me know.
And perl , whose helping you with hindi ?
mujhe pratham puraskaar milne kii aashaa hai , medal?
ohk-now u r indian - and dont lie abt dt ! gt ya !
wt ok ???
Mera naam vivckenanda hai
anjali, if f(x) is continuous at x = 0, then lim f(x) = f(0) as x ->0
anyways, plzzzz ans my maths query , i gtg , cya ltrz !
the answer , he is right
First let's write over g(x) = x * f(x) as g(x) = h(x) * f(x), where h(x) = x We know that h(x) = x is continuous at x = 0. And we know that f(x) is continuous at x = 0 because that is given. We know that the product of continuous functions at a point is continuous at the point (theorem on continuity). So g(x) is continuous at the point. how do we deduce that g(x) is differentiable?
Limit[ (g(x)-g(0))/(x-0), x->0]= Limit[ (xf(x)-0)/(x-0), x->0]=Limit[ (xf(x) )/x, x->0]=Limit[ f(x), x->0]=f(0) From the definition, g is differentiable at - and g'(0)=f(0)
From the definition, g is differentiable at 0 and g'(0)=f(0)
what definition of derivative you used >
you used lim f(x) - f(a) / ( x - a )
seems alright to me
If Limit[ (g(x) -g(b))/(x-b), x->b]=L exists and finite then g is differentiable at x =b and g'(b)=L
Ok let me spruce it up a bit. g is differntiable at x = 0 if ... ok right lim g(x) - g(a) ) / ( x - a) exists ,
if the limit exists then it must necessarily be finite
ok so you demonstrated that lim [ g(x) - g(0) ] / ( x - 0 ) exists as x-> 0 . the limit of that is the same as the limit of f(x) as x->0 , and we know that exists since f is continuous
Some people may consider Infinity as a limit too. So we have to avoid this possibility.
so you demonstrated that lim [ g(x) - g(0) ] / ( x - 0 ) exists as x-> 0 . the limit of that is the same as the limit of f(x) as x->0 , and we know that exists since f is continuous The answer is yes.
oh , good point
but we didnt actually ask for what g'(0) is, so that was extra information
actually you can do this proof with a weaker condition, f(x) limit exists at x = 0