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Can sage help determine if $$ |f(x) - L | < \epsilon$$ is true?

Here's what I want to check:

Given $f(x),\epsilon>0, L\in \mathbb{R}$ and $ N(\epsilon)$

is $$n>N(\epsilon) \Rightarrow |f(n)-L | < 3$$

true?

I thought I could accomplish this using symbolic expressions and assume(). This is what I've tried:

forget()
var('ep, n')
f(x)=1/(n+7)
N = (1/ep)-7


assume(ep>0)
assume(n>N)

show(abs(f(n)-0)<ep)
bool(abs(f(n)-0)<ep)

But the result of is False. What is the proper way to do this in Sage?

Can sage help determine if $$ |f(x) $|f(x) - L | L| < \epsilon$$ \epsilon$ is true?

Here's what I want to check:

Given $f(x),\epsilon>0, L\in \mathbb{R}$ and $ N(\epsilon)$

is $$n>N(\epsilon) \Rightarrow |f(n)-L | < 3$$

true?

I thought I could accomplish this using symbolic expressions and assume(). This is what I've tried:

forget()
var('ep, n')
f(x)=1/(n+7)
N = (1/ep)-7


assume(ep>0)
assume(n>N)

show(abs(f(n)-0)<ep)
bool(abs(f(n)-0)<ep)

But the result of is False. What is the proper way to do this in Sage?

Can sage help determine if $|f(x) - L| < \epsilon$ is true?

Here's what I want to check:

Given $f(x),\epsilon>0, L\in $f(x)$, $\epsilon>0$, $L\in \mathbb{R}$ and $ N(\epsilon)$

is $N(\epsilon)$, is $$n>N(\epsilon) \Rightarrow |f(n)-L | < 3$$

\epsilon$$ true?

I thought I could accomplish this using symbolic expressions and assume(). This is what I've tried:

forget()
var('ep, n')
f(x)=1/(n+7)
N = (1/ep)-7

 assume(ep>0)
assume(n>N)

show(abs(f(n)-0)<ep)
bool(abs(f(n)-0)<ep)

But the result of is False. False. What is the proper way to do this in Sage?