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  • 2.4E: Exercises for Section 2.3
    https://stats.libretexts.org/Courses/Lake_Tahoe_Community_College/Interactive_Calculus_Q1/02%3A_Limits/2.04%3A_The_Limit_Laws/2.4E%3A_Exercises_for_Section_2.3
    then, limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=limx→4(x+4)=4+4=8 then, \(\displaystyle \lim_{x→ 1/2}\frac{2x^2+3x−2}{2x−1}=\lim_{x→1/2}\frac{(2x−1)(x+2)}...then, limx→4x2−16x−4=limx→4(x+4)(x−4)x−4=limx→4(x+4)=4+4=8 then, limx→1/22x2+3x−22x−1=limx→1/2(2x−1)(x+2)2x−1=limx→1/2(x+2)=12+2=52 limx→−52+g(x)f(x)=2+(limx→−5g(x))limx→−5f(x)=2+02=1

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