Jawab:
12. a. 12
13. e. ⅓√7
Penjelasan dengan langkah-langkah:
» No. 12
[tex]\begin{aligned}{\sf{\lim_{x\to6}\frac{12-2x}{3-\sqrt{x+3}}}}&={\sf{\lim_{x\to6}\frac{12-2x}{3-\sqrt{x+3}}\times\frac{3+\sqrt{x+3}}{3+\sqrt{x+3}}}}\\&={\sf{\lim_{x\to6}\frac{(12-2x)(3+\sqrt{x+3})}{(3-\sqrt{x+3})(3+\sqrt{x+3})}}}\\&={\sf{\lim_{x\to6}\frac{2\cancel{(6-x)}(3+\sqrt{x+3})}{\cancel{6-x}}}}\\&={\sf{\lim_{x\to6}2(3+\sqrt{x+3})}}\\&={\sf{2(3+\sqrt{6+3})}}\\&={\sf{2(3+\sqrt{9})}}\\&={\sf{2(3+3)}}\\&={\sf{2(6)}}\\&={\sf{12}}\end{aligned}[/tex]
» No. 13
[tex]\begin{aligned}{\sf{\lim_{x\to3}\frac{x-3}{\sqrt{x^2-2}-\sqrt{7}}}}&={\sf{\lim_{x\to3}\frac{x-3}{\sqrt{x^2-2}-\sqrt{7}}}\times\frac{\sqrt{x^2-2}+\sqrt{7}}{\sqrt{x^2-2}+\sqrt{7}}}\\&={\sf{\lim_{x\to3}\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{(\sqrt{x^2-2}-\sqrt{7})(\sqrt{x^2-2}+\sqrt{7})}}}\\&={\sf{\lim_{x\to3}\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{x^2-9}}}\\&={\sf{\lim_{x\to3}\frac{\cancel{(x-3)}(\sqrt{x^2-2}+\sqrt{7})}{\cancel{(x-3)}(x+3)}}}\\&={\sf{\lim_{x\to3}\frac{\sqrt{x^2-2}+\sqrt{7}}{x+3}}}\\&={\sf{\frac{\sqrt{3^2-2}+\sqrt{7}}{3+3}}}\\&={\sf{\frac{\sqrt{9-2}+\sqrt{7}}{6}}}\\&={\sf{\frac{\sqrt{7}+\sqrt{7}}{6}}}\\&={\sf{\frac{\cancel2\sqrt{7}}{\cancel6}}}\\&={\sf{\frac{1}{3}\sqrt{7}}}\end{aligned}[/tex]
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Jawab:
12. a. 12
13. e. ⅓√7
Penjelasan dengan langkah-langkah:
» No. 12
[tex]\begin{aligned}{\sf{\lim_{x\to6}\frac{12-2x}{3-\sqrt{x+3}}}}&={\sf{\lim_{x\to6}\frac{12-2x}{3-\sqrt{x+3}}\times\frac{3+\sqrt{x+3}}{3+\sqrt{x+3}}}}\\&={\sf{\lim_{x\to6}\frac{(12-2x)(3+\sqrt{x+3})}{(3-\sqrt{x+3})(3+\sqrt{x+3})}}}\\&={\sf{\lim_{x\to6}\frac{2\cancel{(6-x)}(3+\sqrt{x+3})}{\cancel{6-x}}}}\\&={\sf{\lim_{x\to6}2(3+\sqrt{x+3})}}\\&={\sf{2(3+\sqrt{6+3})}}\\&={\sf{2(3+\sqrt{9})}}\\&={\sf{2(3+3)}}\\&={\sf{2(6)}}\\&={\sf{12}}\end{aligned}[/tex]
» No. 13
[tex]\begin{aligned}{\sf{\lim_{x\to3}\frac{x-3}{\sqrt{x^2-2}-\sqrt{7}}}}&={\sf{\lim_{x\to3}\frac{x-3}{\sqrt{x^2-2}-\sqrt{7}}}\times\frac{\sqrt{x^2-2}+\sqrt{7}}{\sqrt{x^2-2}+\sqrt{7}}}\\&={\sf{\lim_{x\to3}\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{(\sqrt{x^2-2}-\sqrt{7})(\sqrt{x^2-2}+\sqrt{7})}}}\\&={\sf{\lim_{x\to3}\frac{(x-3)(\sqrt{x^2-2}+\sqrt{7})}{x^2-9}}}\\&={\sf{\lim_{x\to3}\frac{\cancel{(x-3)}(\sqrt{x^2-2}+\sqrt{7})}{\cancel{(x-3)}(x+3)}}}\\&={\sf{\lim_{x\to3}\frac{\sqrt{x^2-2}+\sqrt{7}}{x+3}}}\\&={\sf{\frac{\sqrt{3^2-2}+\sqrt{7}}{3+3}}}\\&={\sf{\frac{\sqrt{9-2}+\sqrt{7}}{6}}}\\&={\sf{\frac{\sqrt{7}+\sqrt{7}}{6}}}\\&={\sf{\frac{\cancel2\sqrt{7}}{\cancel6}}}\\&={\sf{\frac{1}{3}\sqrt{7}}}\end{aligned}[/tex]