Limit (x -- infinity) (2x square + 7x + 5)/(4 x square + 3x - 1) - Maths - Limits and Derivatives

Question

Limit (x --> infinity) (2x square + 7x + 5)/(4 x square + 3x -
1)

Sparsh Aggarwal · Accepted Answer

Consider
limx→∞2x2+7x+54x2+3x-1=limx→∞
x22+7x+5x2x24+3x-1x2=limx→∞2+7x+5x24+3x-1x2=2 +
7∞ + 5∞4 + 3∞ - 1∞=24=12

Limit (x --> infinity) (2x square + 7x + 5)/(4 x square + 3x - 1)