How do you find the limit of #((h-1)^2+1)/h# as h approaches 0?
2 Answers
Explanation:
Plugging in
#lim_(h->0) ((h-1)^2+1)/h = ((0-1)^2+1)/0=2/0#
Since
The limit is divergent.
Explanation:
The whole point of evaluating a limit is that we do not generally look (or care) at the behaviour when
In this case we have:
# lim_(h rarr 0) ((h-1)^2+1)/h = lim_(h rarr 0) ((h^2-2h+1)+1)/h #
# " " = lim_(h rarr 0) (h^2-2h+2)/h #
# " " = lim_(h rarr 0) (h-2+2/h) #
# " " = lim_(h rarr 0) h - lim_(h rarr 0)2+2lim_(h rarr 0)1/h #
The First limit exists:
# lim_(h rarr 0) h = 0#
The Second limit exists
# lim_(h rarr 0)2 = 2 #
However the third limit diverges
# 2lim_(h rarr 0)1/h rarr oo#
As
Hence the initial limit is divergent.
