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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?

 2 No.2 Revision slelievre 10188 ●9 ●107 ●204 http://carva.org/samue...

### 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?  3 No.3 Revision slelievre 10188 ●9 ●107 ●204 http://carva.org/samue... ### 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?