# Thread: Why is this limit proof wrong?

1. ## Why is this limit proof wrong?

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!

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..

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