## Question

###### What are common mistakes students make when finding the limit of an infinite sequence?

## Answers

#color(red)(oo/oo=1)# and#color(blue)(oo-oo=0)# are typical mistakes. When you think of limitation, be sure to concider the circumstances.## Explanation:

When you think about the limitation,

#lim_(n->oo) a_n# , don't forget that

#oo# isNOTa infinitely large#color(blue)"number"# ,BUTa#color(red)"situation"# #n# is increasing infinitely.[

#color(red)"example1"# ] Where does#n^2/n# go when#n# goes to infinity?Here we must evaluate

#lim_(n->oo)n^2/n#

It is true#lim_(n->oo)n=oo# and#lim_(n->oo)n^2=oo# , but if you think

#lim_(n->oo)n^2/n=oo/oo=1# you arewrong.The situation shows

#n^2# is growing much faster than#n# and thus the limitation cannot be#1# .To solve this, you first need to

#color(red)"reduce the fraction"# and then let#n->oo# .

#lim_(n->oo)n^2/n=color(red)(lim_(n->oo)n/1=lim_(n->oo)n=oo)#

is the correct way.[

#color(blue)"example2"# ] What is#lim_(n->oo)(n-sqrt(n^2+2n))# ?You might think:

#lim_(n->oo)n=oo# and#lim_(n->oo)sqrt(n^2+2n)=oo# and#oo-oo=0# ・・・Just wait! This is wrong again.Here we should

#color(blue)"ratonalize"# #n-sqrt(n^2+2n)# .

#lim_(n->oo)(n-sqrt(n^2+2n))=lim_(n->oo)((n-sqrt(n^2+2n))(n+sqrt(n^2+2n)))/(n+sqrt(n^2+2n))#

#=color(red)(lim_(n->oo)(-2n)/(n+sqrt(n^2+2n)))# ・・(A)(A) is

#oo/oo# again, so divide both numerator and denominator by#n# .(A)

#=color(blue)(-lim_(n->oo)2/(1+sqrt(1+2/n)))#

#=-2/(1+1)=-1# and this is the goal.In general,

#oo/oo# ,#0/0# ,#ooxx0# ,#oo-oo# ,#1^oo# or#oo^0# typed limitations are called#color(green)"indeterminate form"# and they cannot be calculated like numbers. They should be treated carefully.