# prove function is continuous

• Apr 9th 2010, 10:36 AM
sfspitfire23
prove function is continuous
If $f(x) = 5x - 6$, prove that f is continuous in its domain.

So,

Pick any http://www.mathcs.org/analysis/reals/symbols/epsi.gif > 0. Take any sequence { xn } converging to c. Then there exists an integer N such that
|f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<http://www.mathcs.org/analysis/reals/symbols/epsi.gif

My question is, why do we have |xn - c | < http://www.mathcs.org/analysis/reals/symbols/epsi.gif /5? Why not |xn - c | < http://www.mathcs.org/analysis/reals/symbols/epsi.gif? Where did the 5 come from?

Thanks.
• Apr 9th 2010, 10:39 AM
Drexel28
Quote:

Originally Posted by sfspitfire23
If $f(x) = 5x - 6$, prove that f is continuous in its domain.

So,

Pick any http://www.mathcs.org/analysis/reals/symbols/epsi.gif > 0. Take any sequence { xn } converging to c. Then there exists an integer N such that
|f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<http://www.mathcs.org/analysis/reals/symbols/epsi.gif

My question is, why do we have |xn - c | < http://www.mathcs.org/analysis/reals/symbols/epsi.gif /5? Why not |xn - c | < http://www.mathcs.org/analysis/reals/symbols/epsi.gif? Where did the 5 come from?

Thanks.

To counteract the coefficient of the x term?
• Apr 9th 2010, 10:41 AM
sfspitfire23
I guess my question is how do you choose delta in continuous proofs when you dont know what the actual limit is?
• Apr 9th 2010, 10:42 AM
Chandru1
See, the definition of continuity. the delta basically depends on the choice of epsilon. Moreover this function is uniformly continuous.

you have f(x)=5x+6. AS per your proof we have $x_{n} \to c \ \Rightarrow f(x_n) \to f(c)$. Here u have the delta value as epsilon/5.
• Apr 9th 2010, 10:42 AM
Drexel28
Quote:

Originally Posted by sfspitfire23
I guess my question is how do you choose delta in continuous proofs when you dont know what the actual limit is?

You don't in theory. But, I think you know what the limit should be.

It's like induction. You can't use it until you know the answer, but usually the answer is a) apparent or b) unseeable.
• Apr 9th 2010, 10:48 AM
sfspitfire23
So....really.....I should perform the |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<http://www.mathcs.org/analysis/reals/symbols/epsi.gif calculation first and then see what i need to make |xn-c| less than to satisfy the |f(xn)-f(c)| calculation?

This seems odd to me.
• Apr 9th 2010, 10:48 AM
Drexel28
Quote:

Originally Posted by sfspitfire23
So....really.....I should perform the |f(xn)-(5 c - 6)| = |5xn-6-5c+6| = 5|xn-c|<http://www.mathcs.org/analysis/reals/symbols/epsi.gif calculation first and then see what i need to make |xn-c| less than to satisfy the |f(xn)-f(c)| calculation?

This seems odd to me.

Yes.
• Apr 9th 2010, 10:49 AM
Chandru1
you have f(x)=5x+6. AS per your proof we have $x_{n} \to c \ \Rightarrow f(x_n) \to f(c)$. Here u have the delta value as epsilon/5.