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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, \displaystyle \lim_{x→4}\frac{x^2−16}{x−4}= \lim_{x→4}\frac{(x+4)(x−4)}{x−4}=\lim_{x→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, \displaystyle \lim_{x→4}\frac{x^2−16}{x−4}= \lim_{x→4}\frac{(x+4)(x−4)}{x−4}=\lim_{x→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)}{2x−1}=\lim_{x→1/2}(x+2)=\frac{1}{2}+2=\frac{5}{2} \displaystyle \lim_{x→−5}\frac{2+g(x)}{f(x)}=\frac{2+\left(\displaystyle \lim_{x→−5}g(x)\right)}{\displaystyle \lim_{x→−5}f(x)}=\frac{2+0}{2}=1

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