A Sequence is an ordered list of numbers. Each number in a sequence is called a term. For instance, in the sequence 1, 3, 6,9,…., 1 is the first term, 3 is the second term and so on. The notation a1,a2, a3…an is used to denote the different terms in the sequence. A Limit of a Sequence can be defined as, if the terms of a sequence {an} approaches a number L, as n increases, then lim-(n->infinity) a(n)-> = L. Sequence that have limit converge and the sequence that do not have limits diverge. Let us find the limit of the sequence converges or diverges, given an= (3n-2)/n-1. Let us take f(x) = 3x-2/x-1. In finding the limit of a sequence, we write it as lim(x->infinity) [(3x-2)/(x-1)] which gives us 3, and hence lim (n->infinity)an=3 and hence the sequence converges.
So, the limit of a sequence (an) is L given any e >0, there is an N>0 such that modulus of (an-L) is less than e for all n>N. Using this definition finding limits of sequences is simple. Let us consider Limits of Sequences Examples for a better understanding of the given definition.
Example: Find the limit of the Sequence given by a(n)= 1/n^2. Let X(n)=1/n. Then a(n)= X(n). X(n). we know lim(n->infinity) X(n)=0, we get lim-(n->infinity) a(n) = lim-(n->infinity) X(n) .lim-(n->infinity) X(n) =0.0 =0. By induction we can generalize and show that lim(n->infinity) 1/np=0 for any natural number p>0
Example: Find the limit of the Sequence X(n) = (n^2+3n)/(3+n^2)
Solution: X(n) = [1+3n/n^2]/[(3/n^2)+(n^2/n^2) = [1+3/n]/[(3/n^2) +1]
Now applying the limits, we get, lim(n->8) [1+(3/n)]= lim(n->8)1 + lim(n->8)(3/n) = 1+0=1
Also, lim(n->8)[(3/n^2)+1] = lim(n->8)(3/n^2) + lim(n->8)1 = 0+1 = 1
Hence lim(n->8) X(n) = lim(n->8)[1+3/n]/ lim(n->8)[(3/n^2) +1] = 1/1 = 1
Consider the function, f(x)= x^2 sin(1/x). When x gets closer to zero, the function g(x) = sin(1/x) has no limits. But this function g(x) = sin(1/x) is bounded shown as, -1 = sin(1/x) = 1 for any real number x. Since x^2=0, we get –x^2= x^2 sin(1/x) = x^2. Hence when x get closer to 0, x^2 and –x^2 become very small in magnitude. We have, lim(n->0) x^2sin(1/x)=0. This is an example for the Pinching Theorem which is also called the Sandwich Theorem or the Squeeze Theorem, is given as: Let h(x)=f(x) =g(x) for any x is an interval around the point a. If lim(x->a) h(x) = L and lim(x->a) g(x) = L, then lim(x->a) f(x) = L. Basically , the Pinching Theorem says that if a function is trapped between two functions, both of which are approaching a particular value, then the function trapped in between also approaches that same value.
