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GATE2020-CE-1-2
The value of $\displaystyle{} \lim_{x\to\infty}\dfrac{x^2-5x+4}{4x^2+2x}$ is $0 \\$ $\dfrac{1}{4} \\$ $\dfrac{1}{2} \\$ $1$
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$