Jawab:
d. 5
Penjelasan dengan langkah-langkah:
~Pemfaktoran
[tex]\begin{aligned}{\sf{\lim_{x\rightarrow2}\frac{2x^2-3x-2}{x-2}}}&={\sf{\lim_{x\rightarrow2}\frac{(2x+1)\cancel{(x-2)}}{\cancel{x-2}}}}\\&={\sf{\lim_{x\rightarrow2}2x+1}}\\&={\sf{2(2)+1}}\\&={\sf{4+1}}\\&={\sf{5}}\end{aligned}[/tex]
┄
~L'Hospital
[tex]\begin{aligned}{\sf{\lim_{x\rightarrow2}\frac{2x^2-3x-2}{x-2}}}&={\sf{\lim_{x\rightarrow2}\frac{\frac{d}{dx}(2x^2-3x-2)}{\frac{d}{dx}(x-2)}}}\\&={\sf{\lim_{x\rightarrow2}\frac{4x-3}{1}}}\\&={\sf{\lim_{x\rightarrow2}4x-3}}\\&={\sf{4(2)-3}}\\&={\sf{8-3}}\\&={\sf{5}}\end{aligned}[/tex]
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Jawab:
d. 5
Penjelasan dengan langkah-langkah:
~Pemfaktoran
[tex]\begin{aligned}{\sf{\lim_{x\rightarrow2}\frac{2x^2-3x-2}{x-2}}}&={\sf{\lim_{x\rightarrow2}\frac{(2x+1)\cancel{(x-2)}}{\cancel{x-2}}}}\\&={\sf{\lim_{x\rightarrow2}2x+1}}\\&={\sf{2(2)+1}}\\&={\sf{4+1}}\\&={\sf{5}}\end{aligned}[/tex]
┄
~L'Hospital
[tex]\begin{aligned}{\sf{\lim_{x\rightarrow2}\frac{2x^2-3x-2}{x-2}}}&={\sf{\lim_{x\rightarrow2}\frac{\frac{d}{dx}(2x^2-3x-2)}{\frac{d}{dx}(x-2)}}}\\&={\sf{\lim_{x\rightarrow2}\frac{4x-3}{1}}}\\&={\sf{\lim_{x\rightarrow2}4x-3}}\\&={\sf{4(2)-3}}\\&={\sf{8-3}}\\&={\sf{5}}\end{aligned}[/tex]