Question

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

Answer 1

derivable or differentiable? careful with the jargon :P

Answer 2

derivable ! copied from my buk ! :p

Answer 3

i think both means the same

Answer 4

Thats ok ! Solve d prob ! :p

Answer 5

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

Answer 6

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

Answer 7

Woh Hansane wali baat thi anjali

Answer 8

use definition of continuity

Answer 9

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

Answer 10

anjali, mai apne khaane kaa anand le rahaa hoon.

Answer 11

anjali, one minute please

Answer 12

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

Answer 13

Hmmm, ok , search , then let me know.

Answer 14

And perl , whose helping you with hindi ?

Answer 15

:P :D

Answer 16

anjali, hello

Answer 17

heloo

Answer 18

mujhe pratham puraskaar milne kii aashaa hai , medal?

Answer 19

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

Answer 20

:p

Answer 21

ok

Answer 22

wt ok ???

Answer 23

no problem

Answer 24

Mera naam vivckenanda hai

Answer 25

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

Answer 26

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

Answer 27

the answer , he is right

Answer 28

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?

Answer 29

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)

Answer 30

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

Answer 31

what definition of derivative you used >

Answer 32

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

Answer 33

seems alright to me

Answer 34

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

Answer 35

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 ,

Answer 36

if the limit exists then it must necessarily be finite

Answer 37

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

Answer 38

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

Answer 39

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.

Answer 40

oh , good point

Answer 41

about infinity

Answer 42

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

Answer 43

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