MHB Can you check my proof for a limit formula

[Amad27]

Hello,

Suppose there is a $\delta > 0$ such that $f(x) = g(x)$ when $0 < |x - a| < \delta$. Prove that $\displaystyle \lim_{x\to a} f(x) = \lim_{x \to a} g(x)$.

$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$

$|g(x) - M| < \epsilon$ for $|x - a| <\delta_2$

Let $\delta' = \min(\delta_1, \delta_2)$

Let $\delta' < \delta$

Therefore, $f(x) = g(x)$ for this interval $\delta'$

This gives us:

$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$

$|f(x) - M| < \epsilon$ for $|x - a| <\delta_2$

Thus, what is required is to prove $M = L$

Assume $M \ne L$

Since $\epsilon \in (0, \infty)$, this means $\epsilon = |L - M|/2$ is a possibility.

$|f(x) - L| < |L - M|/2$ for $|x - a| < \delta_1$

$|f(x) - M| < |L - M|/2$ for $|x - a| <\delta_2$ $|L - M| = |-(f(x) - L) + (f(x) - M)|$

The triangle inequality states,

$|f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$

Then,

$ |L - M|/2 + |L - M|/2 > |f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$

Which is a contradiction, therefore $L \ne M$ is false, and $L = M$ is the true statement.

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[Fallen Angel]

Hi,

You got the idea but there are some little mistakes, I put it in red

Hello,

Suppose there is a $\delta > 0$ such that $f(x) = g(x)$ when $0 < |x - a| < \delta$. Prove that $\displaystyle \lim_{x\to a} f(x) = \lim_{x \to a} g(x)$.

$|f(x) - L| < \epsilon$ for $|x - a| < \delta_1$
$|g(x) - M| < \epsilon$ for $|x - a| <\delta_2$

Let $\delta' = \min(\delta_1, \delta_2)$

Let $\delta' < \delta$ In essence, we don't let $\delta ' <\delta$, we need to choose $\delta_{1},\delta_{2}$ such that this inequility holds taking $\delta'$ as above

Therefore, $f(x) = g(x)$ for this interval $\delta'$ $\delta '$ is not an interval, but you can skip this sentence.

This gives us:

$|f(x) - L| < \epsilon$ for $|x - a| < $$\delta '$

$|f(x) - M| < \epsilon$ for $|x - a| <$$\delta '$

Thus, what is required is to prove $M = L$

Assume $M \ne L$

Since $\epsilon \in (0, \infty)$, this means $\epsilon = |L - M|/2$ is a possibility.

$|f(x) - L| < |L - M|/2$ for $|x - a| <$$\delta_{3}$

$|f(x) - M| < |L - M|/2$ for $|x - a| <$$\delta_{4}$

Being both $\delta_{3},\delta_{4}<\delta '$

$|L - M| = |-(f(x) - L) + (f(x) - M)|$

The triangle inequality states,

$|f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$

Then,

$ |L - M|/2 + |L - M|/2 > |f(x) - L| + |f(x) - M| \ge |(f(x) - M) - (f(x) - L)|$

Which is a contradiction, therefore $L \ne M$ is false, and $L = M$ is the true statement.

$\space\space\space\space\space\space\space\space\space\space
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\space\space\space\space\space\space\space\space\space\space
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\space\space\space\space\space\space\space\space\space\space
\space\space\space\space\space\space\space\space\space\space \blacksquare$