How do you evaluate the limit of #lim (2n^2)/n# as #n->0#?
1 Answer
Nov 16, 2016
0
Explanation:
evaluating for n = 0 gives indeterminate form.
Simplifying the expression by cancelling.
#lim_(nto0)(2 cancel(n^2)^n)/cancel(n)^1#
#rArrlim_(nto0)2n=0#
