Why is this true? Limit problem... lim x-> infinity (2x^6+5x^4+7x) / (7x^6-3x^5-4x^3+10x) = 2/7?

Question

I cant figure out why this limit is true:
lim x-> infinity (2x^6+5x^4+7x) / (7x^6-3x^5-4x^3+10x) = 2/7

Why? Is L'Hopitals rule somehow involved?

Answer 1

You don't need to use L'Hospital's rule.

Lim x→∞ (2x^6+5x^4+7x) / (7x^6-3x^5-4x^3+10x)

Divide numerator and denominator by x^6.

= Lim x→∞ (2 + 5/x² + 7/x^5) / (7 - 3/x - 4/x³ + 10/x^5)

Notice that as x→∞ all the terms except one in the numerator and one in the denominator go to zero.

= (2 + 0 + 0) / (7 - 0 - 0 + 0) = 2/7

Answer 2

lim(x->-infinity) (5 x^3+10 x)/(7 x-8 x^2)= using L'wellbeing facility's rule we've, lim(x->-infinity) (5 x^3+10 x)/(7 x-8 x^2) = lim(x->-infinity) (( d(10 x+5 x^3))/( dx))/(( d(7 x-8 x^2))/( dx)): = lim(x->-infinity) (15 x^2+10)/(7-16 x) using L'wellbeing facility's rule we've, lim(x->-infinity) (15 x^2+10)/(7-16 x) = lim(x->-infinity) (( d(10+15 x^2))/( dx))/(( d(7-16 x))/( dx)): = lim(x->-infinity) -(15 x)/8 element out constants: = -15/8 (lim(x->-infinity) x) The reduce of x as x approaches -infinity is -infinity: = infinity <--- very last answer wish this helps! :)