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7.2 Q.I · Q22

Q.lim⁡x→−3[x+3x2+4x+3]\displaystyle\lim_{x\to -3}\left[\frac{x+3}{x^2+4x+3}\right]

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✓ Free question

x2+4x+3=(x+3)(x+1)x^2+4x+3=(x+3)(x+1). So x+3(x+3)(x+1)=1x+1\dfrac{x+3}{(x+3)(x+1)}=\dfrac{1}{x+1} for x≠−3x\ne-3. At x=−3x=-3: 1−3+1=−12\dfrac{1}{-3+1}=-\dfrac12.

✓Final answer

−12-\dfrac{1}{2}

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