Evaluate #lim _(x ->1) (x^4-1)/(x-1)# ?

Question

14792 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T08:26:22

#+-oo#

If you mean it that way...
#lim_(x->1)(x^4-1/(x-1))=+-oo#

It is #+oo# if you start with #x>1# and #-oo# if you start with #x<1#

Answer 2 · 2018-03-08T08:31:47

4

#lim_(x->1) (x^4-1)/(x-1)#

#=>lim_(x->1) [(x^2)^2-(1^2)^2]/(x-1)#

#=>lim_(x->1) [(x^2-1)(x^2+1)]/(x-1)#

#=>lim_(x->1) [(x^2-1^2)(x^2+1)]/(x-1)#

#=>lim_(x->1) [cancel((x-1))(x+1)(x^2+1)]/cancel(x-1)#

#=>lim_(x->1) (x+1)(x^2+1)#

#=> (1+1)(1^1+1)#

#=> 2(1+1)#

#=> 2(2)#

#=> 4#