lim_(x→∞) (2x^(2)+2x−sin^(2)x)/(3x^(2)−4x+cos^(2)x)

<p>Jawaban : 2/3</p><p>&nbsp;</p><p>Pembahasan :</p><p>Konsep :</p><p>lim_(x→∞) sin x/x = 0</p><p>lim_(x→∞) cos x/x = 0</p><p>&nbsp;</p><p>&nbsp;</p><p>lim_(x→∞) (2x²+2x−sin²x)/(3x²−4x+cos²x)</p><p>= (2(∞)²+2(∞)−sin²(∞))/(3(∞)²−4(∞)+cos²(∞))</p><p>= tak tentu</p><p>&nbsp;</p><p>Karena tak tentu, maka gunakan sifat limit trigonometri tak hingga.</p><p>&nbsp;</p><p>lim_(x→∞) (2x²+2x−sin²x)/(3x²−4x+cos²x)</p><p>= lim_(x→∞) (2x²+2x−sin²x)/(3x²−4x+cos²x) × (1/x²)/(1/x²)</p><p>= lim_(x→∞) (2+2/x−sin²x/x²)/(3−4/x+cos²x/x²)</p><p>= (2+2/∞ − lim_(x→∞) (sinx/x)²)/(3−4/∞ + lim_(x→∞) &nbsp;(cosx/x)²)</p><p>= (2+0 − 0²)/(3−0 + 0²)</p><p>= 2/3</p><p>&nbsp;</p><p>Jadi, lim_(x→∞) (2x²+2x−sin²x)/(3x²−4x+cos²x) = 2/3</p>