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7.7 III · Q94

Q.lim⁡x→∞[(3x2+4)(4x2−6)(5x2+2)4x6+2x4−1]\displaystyle\lim_{x\to \infty}\left[\frac{(3x^2+4)(4x^2-6)(5x^2+2)}{4x^6+2x^4-1}\right]

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The numerator's leading term is 3x2⋅4x2⋅5x2=60x63x^2\cdot4x^2\cdot5x^2=60x^6. The denominator's leading term is 4x64x^6. Dividing by x6x^6 and letting lower-order terms vanish: 604=15\dfrac{60}{4}=15.

✓Final answer

1515

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