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| − | The Sequential Criterion for a Limit of a Function
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| − | We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function f at a cluster point c from A with regards to sequences (an) from A that converge to c.
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| − | Theorem 1 (The Sequential Criterion for a Limit of a Function): Let f:A→R be a function and let c be a cluster point of A. Then limx→cf(x)=L if and only if for all sequences (an) from the domain A where an≠c ∀n∈N and limn→∞an=c then limn→∞f(an)=L.
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| − | Consider a function f that has a limit L when x is close to c. Now consider all sequences (an) from the domain A where these sequences converge to c, that is limn→∞an=c. The Sequential Criterion for a Limit of a Function says that then that as n goes to infinity, the function f evaluated at these an will have its limit go to L.
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| − | For example, consider the function f:R→R defined by the equation f(x)=x, and suppose we wanted to compute limx→0x. We should already know that this limit is zero, that is limx→0x=0. Now consider the sequence (an)=(1n). This sequence (an) is clearly contained in the domain of f. Furthermore, this sequence converges to 0, that is limn→∞1n=0. If all such sequences (an) that converge to 0 have the property that (f(an)) converges to f(0)=0, then we can say that limn→0f(x)=0.
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| − | We will now look at the proof of The Sequential Criterion for a Limit of a Function.
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| − | Proof: ⇒ Suppose that limx→cf(x)=L, and let (an) be a sequence in A such that an≠c ∀n∈N such that limn→∞an=c. We thus want to show that limn→∞f(an)=L.
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| − | Let ϵ>0. We are given that limx→cf(x)=L and so for ϵ>0 there exists a δ>0 such that if x∈A and 0<∣x−c∣<δ then we have that ∣f(x)−L∣<ϵ. Now since δ>0, since we have that limn→∞an=c then there exists an N∈N such that if n≥N then ∣an−c∣<δ. Therefore an∈Vδ(c)∩A.
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| − | Therefore it must be that ∣f(an)−L∣<ϵ, in other words, ∀n≥N we have that ∣f(an)−L∣<ϵ and so limn→∞f(an)=L.
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| − | ⇐ Suppose that for all (an) in A such that an≠c ∀n∈N and limn→∞an=c, we have that limn→∞f(an)=L. We want to show that limx→cf(x)=L.
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| − | Suppose not, in other words, suppose that ∃ϵ0>0 such that ∀δ>0 then ∃xδ∈A∩Vδ(c)∖{c} such that ∣f(xδ)−L∣≥ϵ0. Let δn=1n. Then there exists xδn=an∈A∩Vδn(c)∖{c}, in other words, 0<∣an−c∣<1n and limn→∞an=c. However, ∣f(an)−L∣≥ϵ0 so limn→∞f(an)≠L, a contradiction. Therefore limx→cf(x)=L. ■
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| | ==Resources== | | ==Resources== |
| | * [http://mathonline.wikidot.com/the-sequential-criterion-for-a-limit-of-a-function The Sequential Criterion for a Limit of a Function] | | * [http://mathonline.wikidot.com/the-sequential-criterion-for-a-limit-of-a-function The Sequential Criterion for a Limit of a Function] |
