# Proving Continuity

• May 19th 2008, 12:59 AM
ah-bee
Proving Continuity
so im supposed to prove that x^2-7 is continuous at x=1

f(1) = -6
and now i need to use the formal definition to prove this.
so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"
• May 19th 2008, 01:58 AM
mr fantastic
Quote:

Originally Posted by ah-bee
so im supposed to prove that x^2-7 is continuous at x=1

f(1) = -6
and now i need to use the formal definition to prove this.
so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"

There's an easy way of doing these. I'll have time to show it later.
• May 19th 2008, 02:00 AM
ah-bee
awesome.. cant wait (Evilgrin)
• May 19th 2008, 03:12 AM
mr fantastic
Quote:

Originally Posted by ah-bee
so im supposed to prove that x^2-7 is continuous at x=1

f(1) = -6
and now i need to use the formal definition to prove this.
so suppose |x-1|<delta, we need to deduce |x^2-1|<epsilon

i tried finding a delta that may satisfy this condition by working backwards, eg. trying to go from |x^2-1|<epsilon to |x-1|<'something'
and then letting delta equal that 'something' (note that working backwards is just a scrap piece of working, once i obtain the delta, i will say, suppose delta =.... and then continue with the proof). is there an easier approach to this or what, coz i dont wanna keep working backwards, which will be hard if i get a nasty function. or i could guess what delta is.. which is still a pain unless there is a systematic way of "guessing"

1. $|(x^2 - 7) - (-6)| = |x^2 - 1| = |x - 1| |x + 1|$.

2. Note that $|x - 1| < \delta_0 \Rightarrow 1 - \delta_0 < x < 1 + \delta_0$.
Only values of x in the neighbourhood of x = 1 are of interest so let ${\color{red}\delta_0 = \frac{1}{2}}$.
Then ${\color{red}\frac{1}{2} < x < \frac{3}{2} \Rightarrow |x - 1| < \frac{1}{2}}$.

3. Under this restricted set of values of x, $|x + 1| \leq \frac{5}{2} < 3$.

4. Therefore $|x - 1| |x + 1| < 3 |x - 1|$ for $|x - 1| < \frac{1}{2}$.

Note that $|x - 1| |x + 1| < \epsilon \Rightarrow 3 |x - 1| < \epsilon \Rightarrow |x - 1| < \frac{\epsilon}{3}$.

5. Let $\delta(\epsilon) = min \left \{ \frac{1}{2}, ~ \frac{\epsilon}{3}\right \}$.

Then for all $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that $0 < |x - 1| < \delta(\epsilon) \Rightarrow |(x^2 - 7) - (-6)| < \epsilon$.

Therefore $\lim_{x \rightarrow 1} (x^2 - 7) = -6$.

The key to freedom is steps 2-3 where you make a convenient restriction on the values of x to get an inequality involving only |x - 1| .....
• May 19th 2008, 04:17 AM
ah-bee
thats great, thanks! but i was told by my lecturer to approach this type of question by supposing delta = something, and then deducing the |f(x)-f(a)|<epsilon. i dunno, might just be some technicalities but your method looks great to me, mainly because there isnt a delta that i can choose straight up and then do some algebraic manipulation to end up with |f(x)-f(a)|<epsilon.
• May 19th 2008, 04:27 AM
mr fantastic
Quote:

Originally Posted by ah-bee
thats great, thanks! but i was told by my lecturer to approach this type of question by supposing delta = something, and then deducing the |f(x)-f(a)|<epsilon. i dunno, might just be some technicalities but your method looks great to me, mainly because there isnt a delta that i can choose straight up and then do some algebraic manipulation to end up with |f(x)-f(a)|<epsilon.

If it survives 48 hours without major criticism from ThePerfectHacker, then you can take it to the bank .....