0 votes 0 votes The value of $\displaystyle{} \lim_{x\to\infty}\dfrac{x^2-5x+4}{4x^2+2x}$ is $0 \\$ $\dfrac{1}{4} \\$ $\dfrac{1}{2} \\$ $1$ Calculus gate2020-ce-1 calculus limits + – go_editor asked Feb 27, 2020 • recategorized Mar 12, 2021 by Lakshman Bhaiya go_editor 5.3k points answer See all 0 reply
0 votes 0 votes divide by x^2 on numerator and denominator. you will get ¼ option B shashankrustagi answered Feb 15, 2021 shashankrustagi 140 points comment Share See all 0 reply Please log in or register to add a comment.