0000:00:00:00:00Start15Number 15ExplanationThe value of limx→3(x+6)tan(2x−6)x2−x−6\lim_{x \to 3} \frac{(x + 6) \tan(2x - 6)}{x^2 - x - 6}limx→3x2−x−6(x+6)tan(2x−6) is....−185-\frac{18}{5}−518−95-\frac{9}{5}−5995\frac{9}{5}59185\frac{18}{5}518275\frac{27}{5}527ExplanationPerform factorization on the denominator limx→3(x+6)tan(2x−6)x2−x−6\lim_{x \to 3} \frac{(x + 6) \tan(2x - 6)}{x^2 - x - 6}x→3limx2−x−6(x+6)tan(2x−6)=limx→3(x+6)tan2(x−3)(x+2)(x−3)= \lim_{x \to 3} \frac{(x + 6) \tan 2(x - 3)}{(x + 2)(x - 3)}=x→3lim(x+2)(x−3)(x+6)tan2(x−3)=limx→3x+6x+2⋅tan2(x−3)x−3= \lim_{x \to 3} \frac{x + 6}{x + 2} \cdot \frac{\tan 2(x - 3)}{x - 3}=x→3limx+2x+6⋅x−3tan2(x−3) Because x→3x \to 3x→3, then x−3→0x - 3 \to 0x−3→0. Using limit property limu→0tanauu=a\lim_{u \to 0} \frac{\tan au}{u} = alimu→0utanau=a =3+63+2⋅21= \frac{3 + 6}{3 + 2} \cdot \frac{2}{1}=3+23+6⋅12=95⋅2= \frac{9}{5} \cdot 2=59⋅2=185= \frac{18}{5}=518
15Number 15ExplanationThe value of limx→3(x+6)tan(2x−6)x2−x−6\lim_{x \to 3} \frac{(x + 6) \tan(2x - 6)}{x^2 - x - 6}limx→3x2−x−6(x+6)tan(2x−6) is....−185-\frac{18}{5}−518−95-\frac{9}{5}−5995\frac{9}{5}59185\frac{18}{5}518275\frac{27}{5}527ExplanationPerform factorization on the denominator limx→3(x+6)tan(2x−6)x2−x−6\lim_{x \to 3} \frac{(x + 6) \tan(2x - 6)}{x^2 - x - 6}x→3limx2−x−6(x+6)tan(2x−6)=limx→3(x+6)tan2(x−3)(x+2)(x−3)= \lim_{x \to 3} \frac{(x + 6) \tan 2(x - 3)}{(x + 2)(x - 3)}=x→3lim(x+2)(x−3)(x+6)tan2(x−3)=limx→3x+6x+2⋅tan2(x−3)x−3= \lim_{x \to 3} \frac{x + 6}{x + 2} \cdot \frac{\tan 2(x - 3)}{x - 3}=x→3limx+2x+6⋅x−3tan2(x−3) Because x→3x \to 3x→3, then x−3→0x - 3 \to 0x−3→0. Using limit property limu→0tanauu=a\lim_{u \to 0} \frac{\tan au}{u} = alimu→0utanau=a =3+63+2⋅21= \frac{3 + 6}{3 + 2} \cdot \frac{2}{1}=3+23+6⋅12=95⋅2= \frac{9}{5} \cdot 2=59⋅2=185= \frac{18}{5}=518
