If exist find the limits. Without using L'Hopital's rule.

Question

Without using L Hopital.

First one:

Lim (2x+12)/(|x+6|)

x->-6

Second one:

Lim ((x²+9)^(0.5)-5)/(x+4)

x->-4.

0.5 means square root

Answer 1

when x < -6, x + 6 is negative. So, l x + 6 l = -(x + 6)

when x > -6, x + 6 is positive. So, l x + 6 l = x + 6

Take "one'sided" limits:

lim_x→--6^- [(2x + 12) / l x+6l] = lim_x→-6^- [2(x+6) / -(x+6)] = -2

lim_x→-6⁺ [(2x+12) / lx + 6l] = lim_x→-6⁺ [2(x+6) / (x+6)] = 2

Since the one-sided limits are unequal, lim_x→-6 [(2x+12) / lx+6l] does not exist.

Answer 2

In the 1st one divide out the factor (x+6), so that the limit is 2.

In the 2nd one, multiply the numerator and denominator by sqrt(x²+9)+5 to get x²-16 in the numerator, then divide out the factor x+4 which leaves x-4 in the numerator and sqrt(x²+9)+5 in the denominator so that the limit is -.8.