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7.2 Q.I · Q23

Q.lim⁡y→0[5y3+8y23y4−16y2]\displaystyle\lim_{y\to 0}\left[\frac{5y^3+8y^2}{3y^4-16y^2}\right]

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5y3+8y2=y2(5y+8)5y^3+8y^2=y^2(5y+8) and 3y4−16y2=y2(3y2−16)3y^4-16y^2=y^2(3y^2-16). Cancelling y2y^2 (valid since y→0y\to0 but y≠0y\ne0): 5y+83y2−16\dfrac{5y+8}{3y^2-16}. At $y=0 …

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