PART A: Include a graph of the sequence. Include a sentence telling us what the limiting value visually looks like. The value found "graphically" should match the L value that was found for their limit.If there is a discrepancy, explore where this may be coming from and work to fix the error.Problem:bn=+2nFind the limit for bnasn approaches .limn(+2n)=2n approaches 0, hence the limitis>0be given. We need to find an integer NinN such that when nN,then |+2n-|<.Since n>2,we conclude that N>2(see the scratchwork for n and N below).Since nN,we have the following:i|+2n-|=|2n|=2n,iinN, then 2n2(N)<222 and N>2PART B: Create a table of values. Suggested values include n=1,10,100 and 1000. Show that the limit trends towards the value found algebraically and graphically. Imagine that you are anAP Calculus teacher in a high school Find the limitof the sequence cn=n6n+2.Solving Algebraically: limnn6n+2=16.limnn6n+2=limn16+2n=16as2n approaches 0asn approaches infinity.Proof: Suppose >0. Choose N=N where N>118. Let ninN satisfy nN. Then, nN implies n>118 and therefore 18n+12>1,which is the same as118n+12<. Hence we have |n6n+2-16|<, and by the definition of the limitof a sequence, limnn6n+2=16. QEDScratch Work:|n6n+2-16|=|6n6(6n+2)-6n+26(6n+2)|=|-26(6n+2)|=118n+12<118n,(asn1)<118N,< for all nN118.