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NCERT Class XI Mathematics - Limits and Derivatives - Solutions

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Question : 8 of 72
Marks: +1, -0
lim⁡x→3x4−812x2−5x−3\lim\limits_{x\to 3}\frac{x^4-81}{2x^2-5x-3}
Solution:  
We have, lim⁡x→3x4−812x2−5x−3\lim\limits_{x\to 3}\frac{x^4-81}{2x^2-5x-3} (0/0 form) = lim⁡x→3(x2−9)(x2+9)2x2−6x+x−3\lim\limits_{x\to 3}\frac{(x^2-9)(x^2+9)}{2x^2-6x+x-3}
= lim⁡x→3(x−3)(x+3)(x2+9)2x(x−3)+1(x−3)\lim\limits_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{2x(x-3)+1(x-3)} = lim⁡x→3(x−3)(x+3)(x2+9)(2x+1)(x−3)\lim\limits_{x\to 3}\frac{(x-3)(x+3)(x^2+9)}{(2x+1)(x-3)}
= (3+3)(9+9)6+1\frac{(3+3)(9+9)}{6+1} = 6×187\frac{6\times 18}{7} = 1087\frac{108}{7}.
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