How do you find #\lim _ { x \rightarrow + \infty } \sqrt { 16x ^ { 2} + 8x + 6} - \sqrt { 16x ^ { 2} - 8x - 6}#?

1 Answer
Jim H
Jul 15, 2017

Please see below.

Explanation:

#lim_(xrarroo)(sqrt(16x^2+8x+6)-sqrt(16x^2-8x-6)) = lim_(xrarroo)((sqrt(16x^2+8x+6)-sqrt(16x^2-8x-6))/1 * (sqrt(16x^2+8x+6)+sqrt(16x^2-8x-6))/(sqrt(16x^2+8x+6)+sqrt(16x^2-8x-6)))#

# = lim_(xrarroo) ((16x^2+8x+6)-(16x^2-8x-6))/(sqrt(16x^2+8x+6)+sqrt(16x^2-8x-6))#

# = lim_(xrarroo) (16x+12)/(sqrt(16x^2+8x+6)+sqrt(16x^2-8x-6))#

# = lim_(xrarroo) (x(16+12/x))/(sqrt(x^2)(sqrt(16+8/x+6/x^2)+sqrt(16-8/x-6/x^2))#

# = lim_(xrarroo) (x(16+12/x))/(abs(x)(sqrt(16+8/x+6/x^2)+sqrt(16-8/x-6/x^2))#

# = lim_(xrarroo) (16+12/x)/(sqrt(16+8/x+6/x^2)+sqrt(16-8/x-6/x^2)#

# = (16+0)/(sqrt(16+0+0)-sqrt(16+0+0)) = 16/8 = 2#

Bonus

For #x < 0#, we have #sqrt(x^2) = abs(x) = -x#.

So

#lim_(xrarr-oo)(sqrt(16x^2+8x+6) -sqrt(16x^2-8x-6)) = -2#

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