How do you find the limit of #x^5/(4x^7-x^3+9)# as x approaches infinity?

Question

2369 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-04T09:38:37

The limit as #x->oo# is #=0#

There are #2# ways for calculating this limit

First method

We take the term of highest degree in the numerator and in the denominator

#lim_(x->+-oo)(x^5)/(4x^7-x^3+9)=lim_(x->+-oo)(x^5)/(4x^7)#

#=lim_(x->+-oo)(1)/(4x^2)=0#

Second method

Divide the denominator by #x^7#

#(x^5)/(4x^7-x^3+9)=(x^5)/(x^7(4-1/x^4+9/x^7))#

#=1/(x^2(4-1/x^4+9/x^7))#

#lim_(x->+-oo)1/x^4=0#

#lim_(x->+-oo)9/x^7=0#

Therefore,

#lim_(x->+-oo)(x^5)/(4x^7-x^3+9)=lim_(x->+-oo)1/(4x^2)=0#