# How do you find the limit #lim (2x^2-3x+2)/(x^3+5x^2-1)# as #x->oo#?

##### 1 Answer

#### Explanation:

Note that as

We can therefore multiply numerator and denominator by

# lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1) = lim_(x rarr oo) (2x^2-3x+2)/(x^3+5x^2-1)*(1/x^3)/(1/x^3)#

# " " = lim_(x rarr oo) ((1/x^3)(2x^2-3x+2))/((1/x^3)(x^3+5x^2-1))#

# " " = lim_(x rarr oo) (2/x-3/x^2+2/x^3)/(1+5/x-1/x^3)#

# " " = (0-0+0)/(1+0-0)#

# " " = 0#

We can verify this result by looking at the graph of

graph{(2x^2-3x+2)/(x^3+5x^2-1) [-7, 13, -4.16, 5.84]}

and indeed it does appear that for large