What is the limit of #(x^2 + 1) / (2x^2 -3x -2)# as x goes to infinity?

Question

What is the limit of #(x^2 + 1) / (2x^2 -3x -2)# as x goes to infinity?

1 Answer

Impact of this question

4908 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-19T19:01:30

I found #1/2#

Explanation:

You can collect #x^2# to get:
#lim_(x->oo)(x^2(1+1/x^2))/(x^2(2-3/x-2/x^2))=lim_(x->oo)(cancel(x^2)(1+1/x^2))/(cancel(x^2)(2-3/x-2/x^2))=lim_(x->oo)(1+1/x^2)/(2-3/x-2/x^2)=# as #x->oo# then #1/x^n->0# so:
#=(1+0)/(2-0-0)=1/2#

Science

Math

Humanities

... and beyond

What is the limit of #(x^2 + 1) / (2x^2 -3x -2)# as x goes to infinity?

1 Answer

Explanation:

Related questions

Impact of this question