# question on simple calculus proof

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Hi there,

I have been doing some calculus exercises, and was wondering if I had this right.

Prove that $f$ is continuous at $a$ if and only if

$\lim_{h \to 0}f(a+h) = f(a)$

I'm not sure if I got this correct, but I assumed that as we take $h$ to $0$ then $f(a+h) = f(a+0) = f(a)$ and so we are done?

Thanks

Edited by BSZDcZMX
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I'm not sure if I got this correct, but I assumed that as we take $h$ to $0$ then $f(a+h) = f(a+0) = f(a)$ and so we are done?

I don't agree. Your reasoning applies equally well whether f is continuous at a or not.

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Hi there,

I have been doing some calculus exercises, and was wondering if I had this right.

Prove that $f$ is continuous at $a$ if and only if

$\lim_{h \to 0}f(a+h) = f(a)$

I'm not sure if I got this correct, but I assumed that as we take $h$ to $0$ then $f(a+h) = f(a+0) = f(a)$ and so we are done?

Thanks

Often the statement you are trying to prove is the definition of continuity. What is the definition of continuity yiou are given?

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Hi guys,

Thanks for the replies, very helpful. I'm sure it's right now!

mathematic, the definition of continuity is that the function is continuous everywhere on it's domain, and so the limit everywhere on the domain exists, so essentially the question was that I was told the limit was $f(a)$ as the limit approached $0$, and so all I had to do was take $h$ to $0$ in the limit to prove that it was continuous.

I stuggle with the logic, here is a harder proof I attempted before and I was hoping to get some verification on my attempt:

Prove the statement using the $\epsilon , \delta$ definition of a limit:

$\lim_{x \to -2}(x^2 - 1) = 3$

if $0 < |x -(-2)| < \delta$ then $|(x^2 - 1) - 3| < \epsilon$

we then have:

$|(x^2 - 1) - 3| < \epsilon\rightarrow |x^2 -2| < \epsilon\rightarrow |x + 1||x - 1| < \epsilon\rightarrow |x - 1| < - \frac{\epsilon}{|x + 1|}$

this suggests:

$\delta = \frac{\epsilon}{|x + 1|} \Leftrightarrow \epsilon = |x + 1| \delta$

testing our new $\delta$ :

$|(x^2 - 1) -3| < \delta |x + 1|\rightarrow |x + 1||x - 1| < \delta |x + 1|\rightarrow |x - 1| < \delta$

and we have proved the statement.

I have a feeling that the above is fishy because of the fact that I derived the new $\delta$ from the given statement? Hope someone can help out.

Thanks!

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"mathematic, the definition of continuity is that the function is continuous everywhere on it's domain, and so the limit everywhere on the domain exists"

It sounds like the definition is that a function is continuous if it is continuous. Somethings missing!!

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"mathematic, the definition of continuity is that the function is continuous everywhere on it's domain, and so the limit everywhere on the domain exists"

It sounds like the definition is that a function is continuous if it is continuous. Somethings missing!!

Yeah I got it

And the other proof, that alright?

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$|(x^2 - 1) - 3| < \epsilon\rightarrow |x^2 -2| < \epsilon\rightarrow |x + 1||x - 1| < \epsilon\rightarrow |x - 1| < - \frac{\epsilon}{|x + 1|}$

Check your math here. $x^{2}-1-3=x^{2}-2$?

Edited by DJBruce
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Check your math here. $x^{2}-1-3=x^{2}-2$?

Sorry, mistake(s)!

$\lim_{x \to -2} (x^2 - 1) = 3$

if $0 < |x - (-2)| < \delta$ then $|(x^2 - 1) - 3| < \epsilon$

we then have:

$|(x^2 - 1) - 3| < \epsilon \rightarrow|x^2 - 4| < \epsilon \rightarrow|x^2 - 2^2| < \epsilon \rightarrow|x + 2||x - 2| < \epsilon \rightarrow|x + 2| < \frac{\epsilon}{|x - 2|}$

this suggests:

$\delta = \frac{\epsilon}{|x - 2|} \Leftrightarrow \epsilon = |x - 2| \delta$

testing new $\delta$:

$|(x^2 - 1) - 3| < |x - 2| \delta \rightarrow|x + 2||x - 2| < |x - 2| \delta \rightarrow|x + 2| < \delta$

and proved.

Sorry for the error, very careless of me!

EDIT: I can't edit my original post, but the correction is above.

Any feedback as to the proof?

Edited by BSZDcZMX

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