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Miscellaneous 7 II · Q137

Q.lim⁡x→∞(8x2+5x+32x2−7x−5)4x+38x−1\displaystyle\lim_{x\to \infty}\left(\frac{8x^2+5x+3}{2x^2-7x-5}\right)^{\frac{4x+3}{8x-1}}

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As x→∞x\to\infty, dividing numerator and denominator of the base by x2x^2: 8x2+5x+32x2−7x−5→82=4\dfrac{8x^2+5x+3}{2x^2-7x-5}\to\dfrac82=4. The exponent 4x+38x−1→48=12\dfrac{4x+3}{8x-1}\to\dfrac48=\dfrac12 (dividing by xx). Since the base tends to the finite positive number 44 (not 11) and the exponent tends to the finite number $1/2 …

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