First, we'll try to plug in the value: #lim_{x to -oo}x+sqrt(x^2+2x) = -oo + sqrt(oo-oo)# We're already encountering a problem: it is simply not allowed to have #oo-oo#, it's like dividing by zero. We need to try a different approach. Whenever I see this kind of limit, I try to use a trick: #lim_{x to -oo}x+sqrt(x^2+2x)# #= lim_{x to -oo}x+sqrt(x^2+2x)*(x-sqrt(x^2+2x))/(x-sqrt(x^2+2x))# These are the same becaus the factor we're multiplying with is essentially #1#. Why are we doing this? Because there exists a formula which says: #(a-b)(a+b) = a^2-b^2# In this case #a = x# and #b = sqrt(x^2+2x)# Let's apply this formula: #lim_{x to -oo}(x^2-(sqrt(x^2+2x))^2)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(x^2-x^2-2x)/(x-sqrt(x^2+2x))# #= lim_{x to -oo}(-2x)/(x-sqrt(x^2+2x))# Now we're going to use another trick. We'r going to use this one, because we want to get the #x^2# out of the square root: #lim_{x to -oo}(-2x)/(x-sqrt(x^2(1+2/x))# If you look carefully, you see it's the same thing. Now, you might say that #sqrt(x^2) = x#, but you have to remember that #x# is a negative number. Because we're taking the positive square root, #sqrt(x^2) = -x# in this case. #= lim_{x to -oo}(-2x)/(x+xsqrt(1+2/x))# #= lim_{x to -oo}(-2x)/(x(1+sqrt(1+2/x)))# We can cancel the #x#: #= lim_{x to -oo}(-2)/(1+sqrt(1+2/x))# And now, we can finally plug in the value: #= -2/(1+sqrt(1+2/-oo))# A number divided by infinity, is always #0#: #= -2/(1+sqrt(1+0)) = -2/(1+1) = -2/2 = -1# This is the final answer. Hope it helps.
Вакуленко
18.11.2021
1- Найти такое положительное число m чтобы данное выражение было квадратом суммы или разности : 1) x² - 6x + m = x² - 2 * 3 * x + 9 = (х - 3)², m = 9 2) x² + 16x + m = x² + 2 * 8 * x + 64 = (x + 8)², m = 64 3) x² - mx + 9 = x² - 2 * 3 * x + 9 = (x - 3)², m = 6
2. Решить уравнение 1) x² - 3x - 10 = 0 а = 1; b = -3; c = -10 D = b² - 4ac = (-3)² - 4 * 1 * (-10) = 9 + 40 = 49
x+4+x+3=10
(-x+4)+x-3=10
x+4-(x-3)=10
-(x+4)-(x+3)=10
x1=4,5; x2=5,5