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Question 24 of 40

Q.Find the value of lim⁡x→−32x2+7x+33x2+8x−3\lim_{x \to -3} \dfrac{2x^2 + 7x + 3}{3x^2 + 8x - 3}.

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2022Subjective· 4mImportance★★★★★
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Factorise: 2x2+7x+3=(2x+1)(x+3)2x^2+7x+3=(2x+1)(x+3), 3x2+8x−3=(3x−1)(x+3)3x^2+8x-3=(3x-1)(x+3); cancel (x+3)(x+3) and substitute x=−3x=-3 to get 12\tfrac12.

Given: lim⁡x→−32x2+7x+33x2+8x−3\lim_{x\to -3}\dfrac{2x^2+7x+3}{3x^2+8x-3}.

Check the form at x=−3x=-3: numerator =2(9)+7(−3)+3=18−21+3=0=2(9)+7(-3)+3=18-21+3=0; denominator =3(9)+8(−3)−3=27−24−3=0=3(9)+8(-3)-3=27-24-3=0. So it is 00\dfrac{0}{0} — factorise.

Step 1 — factorise.

2x2+7x+3=(2x+1)(x+3),3x2+8x−3=(3x−1)(x+3).2x^2+7x+3=(2x+1)(x+3),\qquad 3x^2+8x-3=(3x-1)(x+3).

Step 2 — cancel (x+3)(x+3). …

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