# continous function

• Jun 13th 2012, 08:22 AM
saravananbs
continous function
f(x)=e-1/x2 if x is not =0
=0 if x=0

then f is continuous. how it can be proved.
• Jun 13th 2012, 08:36 AM
Prove It
Re: continous function
Quote:

Originally Posted by saravananbs
f(x)=e-1/x2 if x is not =0
=0 if x=0

then f is continuous. how it can be proved.

See if \displaystyle \displaystyle \begin{align*} \lim_{x \to 0^-}e^{-\frac{1}{x^2}} = \lim_{x \to 0^+}e^{-\frac{1}{x^2}} = 0 \end{align*}.
• Jun 13th 2012, 08:38 AM
Reckoner
Re: continous function
Quote:

Originally Posted by saravananbs
f(x)=e-1/x2 if x is not =0
=0 if x=0

then f is continuous. how it can be proved.

We need to show that $\displaystyle \lim_{x\to0}f(x) = f(0).$

$\displaystyle \lim_{x\to0}f(x)$

$\displaystyle =\lim_{x\to0}e^{-1/x^2}$

Now evaluate the limit and show that it equals $\displaystyle f(0).$
• Jun 13th 2012, 08:40 AM
saravananbs
Re: continous function
yes at x=0, it is continuous

but for other points , how its derived.

is it like limit x-> a+h f(x) = limit x-> a-h f(x) =f(a)?
• Jun 13th 2012, 08:55 AM
emakarov
Re: continous function
Quote:

Originally Posted by saravananbs
yes at x=0, it is continuous

but for other points , how its derived.

One way is to say that $\displaystyle e^{-1/x^2}$ is continuous as the composition of continuous functions.
• Jun 13th 2012, 09:05 AM
saravananbs
Re: continous function
Quote:

Originally Posted by emakarov
One way is to say that $\displaystyle e^{-1/x^2}$ is continuous as the composition of continuous functions.

i think three function are involved , e^x ,1/x, x^2

all are continuous. is it right.
• Jun 13th 2012, 09:06 AM
Prove It
Re: continous function
All are continuous where x =/= 0.
• Jun 13th 2012, 09:06 AM
emakarov
Re: continous function
Yes, all are continuous for x ≠ 0.