Let g(x)=x f(x) for all x belonging to R
Show that if f is continuous at x=0 then g is derivable at x=0

- anonymous

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- anonymous

derivable or differentiable? careful with the jargon :P

- anonymous

derivable ! copied from my buk ! :p

- experimentX

i think both means the same

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- anonymous

Thats ok ! Solve d prob ! :p

- anonymous

it definitely does not mean the same. if that's a calculus book it's a mistake.

- anonymous

malcom---its derivable ! No prob wd d jargon -solve it plzzz !

- perl

Woh Hansane wali baat thi anjali

- perl

use definition of continuity

- anonymous

I dint gt u ? Ur nt indian rite ? And plz no hints fr ques - straightforward explanatory ans ! I would be really obliged ! :p

- perl

anjali, mai apne khaane kaa anand le rahaa hoon.

- perl

anjali, one minute please

- experimentX

somewhere i had read .. the product of two continuous function is a continuous function is a continuous function ... somehow i can't find it

- anonymous

Hmmm, ok , search , then let me know.

- anonymous

And perl , whose helping you with hindi ?

- anonymous

:P :D

- perl

anjali, hello

- anonymous

heloo

- perl

mujhe pratham puraskaar milne kii aashaa hai , medal?

- anonymous

ohk-now u r indian - and dont lie abt dt ! gt ya !

- anonymous

:p

- perl

ok

- anonymous

wt ok ???

- perl

no problem

- perl

Mera naam vivckenanda hai

- perl

anjali,
if f(x) is continuous at x = 0, then lim f(x) = f(0) as x ->0

- anonymous

anyways, plzzzz ans my maths query , i gtg , cya ltrz !

- perl

the answer , he is right

- perl

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?

- anonymous

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)

- anonymous

From the definition, g is differentiable at 0 and g'(0)=f(0)

- perl

what definition of derivative you used >

- perl

you used
lim f(x) - f(a) / ( x - a )

- experimentX

seems alright to me

- anonymous

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

- perl

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 ,

- perl

if the limit exists then it must necessarily be finite

- perl

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

- anonymous

Some people may consider Infinity as a limit too. So we have to avoid this possibility.

- anonymous

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.

- perl

oh , good point

- perl

about infinity

- perl

but we didnt actually ask for what g'(0) is, so that was extra information

- perl

actually you can do this proof with a weaker condition, f(x) limit exists at x = 0

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