# 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#