Need help with limit proof. (1 Viewer)

Librah · Mar 30, 2015

Having trouble figuring out how to prove this using the epsilon/delta definition, and proofs using epsilon/delta in general.

lim x->0 g(x)=0 where g(x)=3x when x is rational and g(x)=7x when x is irrational.

InteGrand · Mar 30, 2015

Librah said:
Having trouble figuring out how to prove this using the epsilon/delta definition, and proofs using epsilon/delta in general.

lim x->0 g(x)=0 where g(x)=3x when x is rational and g(x)=7x when x is irrational.

We want to show that given any $\varepsilon >0$ , $\exists \delta >0$ such that $|g(x)-0|<\varepsilon \Leftrightarrow |g(x)|<\varepsilon$ whenever $0<|x-0|<\delta \Leftrightarrow 0<|x|<\delta$ .

Clearly $|3x|<|7x|$ for all $x\neq 0$ .

Hence $|g(x)|\leq |7x|$ for all $x\neq 0$ .

So,

$|g(x)|<\varepsilon \Leftarrow |7x|<\varepsilon, x\neq 0$

$\Leftrightarrow |x|<\frac{\varepsilon}{7}$ .

So taking $\delta = \frac{\varepsilon}{7}$ works. Hence the limit is indeed 0.

Librah · Mar 30, 2015

InteGrand said:
We want to show that given any $\varepsilon >0$ , $\exists \delta >0$ such that $|g(x)-0|<\varepsilon \Leftrightarrow |g(x)|<\varepsilon$ whenever $0<|x-0|<\delta \Leftrightarrow 0<|x|<\delta$ .

Clearly $|3x|<|7x|$ for all $x\neq 0$ .

Hence $|g(x)|\leq |7x|$ for all $x\neq 0$ .

So,

$|g(x)|<\varepsilon \Leftarrow |7x|<\varepsilon, x\neq 0$

$\Leftrightarrow |x|<\frac{\varepsilon}{7}$ .

So taking $\delta = \frac{\varepsilon}{7}$ works. Hence the limit is indeed 0.

Doesn't the rational/irrational part have any relevance to the question?

InteGrand · Mar 30, 2015

Librah said:
Doesn't the rational/irrational part have any relevance to the question?

Well, it basically just means $g(x)$ either equals $3x$ or $7x$ . Either way, $|g(x)|\leq|7x|$ .

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Need help with limit proof. (1 Viewer)

Librah

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InteGrand

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Librah

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InteGrand

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