Why is this limit proof wrong?

**Deadstar** · November 20th 2007, 02:40 PM

This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?

Prove using the $\epsilon - \delta$ definition of a limit that $\lim_{x\to 3} 7x = 28$ (was originally 21 of course)

Rough Work:

|f(x) - L| = |7x - 28| = 7|x - 4|

|x - p| we can control with $\delta$

If $|x - 4| < \delta$ , then $|f(x) - L| < 7|x - 4| < 7\delta$

Proof.

Let $\epsilon > 0$ be given. Choose $\delta = \frac{\epsilon}{7}$ , then if $0 < |x - 4| < \delta$ => $|f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

From the $\epsilon - \delta$ definition of a limit, $\lim_{x\to 3} 7x = 28$

Can somebody point out the mistake in that? surely there must be some step thats wrong!

**kalagota** · November 20th 2007, 02:44 PM

Originally Posted by Deadstar

This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?

Prove using the $\epsilon - \delta$ definition of a limit that $\lim_{x\to 3} 7x = 28$ (was originally 21 of course)

Rough Work:

|f(x) - L| = |7x - 28| = 7|x - 4|

|x - p| we can control with $\delta$

If $|x - 4| < \delta$ , then $|f(x) - L| < 7|x - 4| < 7\delta$

Proof.

Let $\epsilon > 0$ be given. Choose $\delta = \frac{\epsilon}{7}$ , then if $0 < |x - 4| < \delta$ => $|f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

From the $\epsilon - \delta$ definition of a limit, $\lim_{x\to 3} 7x = 28$

Can somebody point out the mistake in that? surely there must be some step thats wrong!

if there was no typo error, then it is really wrong..
because x approcahes 3, not 4.. and if it approaches 3, then the limit cannot be 28..

**CaptainBlack** · November 20th 2007, 11:02 PM

Originally Posted by Deadstar

This may be a stupid question thats gonna easily be proved but i just wondered at which step is this wrong. This is taken from a proof in my uni class that limit x->3 (7x) = 21 but just changing one number. why is it wrong and where does it go wrong?

Prove using the $\epsilon - \delta$ definition of a limit that $\lim_{x\to 3} 7x = 28$ (was originally 21 of course)

Rough Work:

|f(x) - L| = |7x - 28| = 7|x - 4|

|x - p| we can control with $\delta$

If $|x - 4| < \delta$ , then $|f(x) - L| < 7|x - 4| < 7\delta$

Proof.

Let $\epsilon > 0$ be given. Choose $\delta = \frac{\epsilon}{7}$ , then if $0 < |x - 4| < \delta$ => $|f(x) - L| < 7\delta = 7\frac{\epsilon}{7} = \epsilon$

From the $\epsilon - \delta$ definition of a limit, $\lim_{x\to 3} 7x = 28$

Can somebody point out the mistake in that? surely there must be some step thats wrong!

Where you take $|x - 4| < \delta$ , means you are constructing the limit as $x \to 4$ not $3$ .

RonL

Thread: Why is this limit proof wrong?

LinkBack

Thread Tools

Display

Why is this limit proof wrong?

Similar Math Help Forum Discussions

What is wrong with this 'limit' proof?

How am I looking at this Limit wrong?

Wrong limit value, why?

Why am I getting this L'Hospital limit wrong?!

limit where am I going wrong?

Search Tags