A limit controls outputs near a point
Objective: Why does the definition exclude x=a even when f(a) is defined?
The statement lim(x→a)f(x)=L concerns values f(x) for inputs close to a but different from a. It does not initially say that f(a) exists or equals L. A graph suggests approach, a table samples it, and algebra may expose it, but the definition asks for a uniform guarantee: every sufficiently close permitted input must force the output into any requested tolerance around L. This separates evidence from proof.
In the ε–δ definition, ε names the desired output accuracy and δ names an input closeness that is sufficient to achieve it. The order matters: an opponent chooses any ε>0, then the proof constructs δ>0 before the input x is tested. The punctured condition 0<|x−a|<δ deliberately omits a. A successful proof shows how output error is bounded by input error without selecting only favorable sample points.
Picture turning an input dial toward a without quite clicking onto a. The limit asks where the output needle can be forced to stay once the input is close enough. It is not a guess from a few readings: after someone names any output error they will tolerate, you must name an input distance that makes every permitted reading obey it.
For f(x)=3x−2, lim(x→a)f(x)=3a−2 for every real a.
Let ε>0 be given. The output error is |(3x−2)−(3a−2)|=3|x−a|.
Choose δ=ε/3. If 0<|x−a|<δ, multiplying by 3 gives 3|x−a|<3δ=ε.
Therefore |f(x)−(3a−2)|<ε for every qualifying x, exactly matching the quantified definition.
Prove directly that lim(x→2)(5x+1)=11, and state the δ that works for a requested ε.
- Compute the error: |(5x+1)−11|=|5x−10|=5|x−2|.
- To make this smaller than ε, it is enough to require |x−2|<ε/5, so choose δ=ε/5.
- Then every x with 0<|x−2|<δ satisfies |(5x+1)−11|<5δ=ε.
Result: δ=ε/5 proves the limit for every ε>0.