lim【x→∞】(【根号】3x^2+8x+6【根号结束】-【根号】3x^2+x+1【根号结束】)
问题描述:
lim【x→∞】(【根号】3x^2+8x+6【根号结束】-【根号】3x^2+x+1【根号结束】)
答
lim(x→∞) [√(3x² + 8x + 6) - √(3x² + x + 1)]
= lim(x→∞) [√(3x² + 8x + 6) - √(3x² + x + 1)] · [√(3x² + 8x + 6) + √(3x² + x + 1)]/[√(3x² + 8x + 6) + √(3x² + x + 1)],乘以共轭数,进行分子有理化
= lim(x→∞) [(3x² + 8x + 6) - (3x² + x + 1)]/[√(3x² + 8x + 6) + √(3x² + x + 1)],(a - b)(a + b) = a² - b²
= lim(x→∞) (7x + 5)/[√(3x² + 8x + 6) + √(3x² + x + 1)]
= lim(x→∞) (7 + 5/x)/[√(3x² + 8x + 6)/x + √(3x² + x + 1)/x],分子分母都除以x
= lim(x→∞) (7 + 5/x)/[√((3x² + 8x + 6)/x²) + √((3x² + x + 1)/x²)]
= lim(x→∞) (7 + 5/x)/[√(3 + 8/x + 6/x²) + √(3 + 1/x + 1/x²)]
= (7 + 0)/[√(3 + 0 + 0) + √(3 + 0 + 0)],当x趋向无穷大时,1/x趋向0
= 7/(√3 + √3)
= 7/(2√3)