negative inifnte limits

Question

Evaluate the limit as x-> -oo (sqrt((x^2)+4x))/3x

Answer 1

Let f(x) = [√(x²+4x)]/ 3x

lim_x->-∞f(x) = lim_x->∞f(-x)

f(-x) = [√((-x)² - 4x)] / 3(-x) = [√(x² - 4x)] / (-3)x

Move x in the denominator under the radical. Remember x-> +∞.

f(-x) = [√(x² - 4x)/x²] / -3 = [√x²/x² - 4/x] / -3

Taking the limit as x - > +∞

[√1/1 - 0) / -3 = 1/-3 or - (1/3)

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