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.
11п/9 = п+(2п/9), п<11п/9, 11п/9 < (3п/2), <=> 11/9<3/2 <=> 11*2 < 3*9 <=> 22< 27, истина. т.о. 11п/9 принадлежит третьей четверти, в которой синус отрицателен, т.е. sin(11п/9) < 0. 3,14<п<3,15. 3,14*(3/2)<(3п/2)<3,15*(3/2)=4,725<5, 5<6,28=2*3,14<2п<2*3,15. (3п/2)<5<2п. Угол в 5 (радиан) принадлежит четвертой четверти, в которой косинус положителен, поэтому cos(5)>0. (3п/2)=1,5п<1,6п<2п. Угол 1,6п принадлежит четвертой четверти, в которой tg отрицателен, т.е. tg(1,6п) <0. ответ. в).
#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.
п<11п/9,
11п/9 < (3п/2), <=> 11/9<3/2 <=> 11*2 < 3*9 <=> 22< 27, истина.
т.о. 11п/9 принадлежит третьей четверти, в которой синус отрицателен, т.е. sin(11п/9) < 0.
3,14<п<3,15.
3,14*(3/2)<(3п/2)<3,15*(3/2)=4,725<5,
5<6,28=2*3,14<2п<2*3,15.
(3п/2)<5<2п.
Угол в 5 (радиан) принадлежит четвертой четверти, в которой косинус положителен, поэтому cos(5)>0.
(3п/2)=1,5п<1,6п<2п.
Угол 1,6п принадлежит четвертой четверти, в которой tg отрицателен, т.е. tg(1,6п) <0.
ответ. в).