What is the limit of # (x +sqrt((x^2)+(3x)))# as x goes to infinity?
2 Answers
Explanation:
# = x +sqrt(x^2)sqrt(1+3/x))# for#x != 0#
# = x +absx sqrt(1+3/x))# for#x != 0#
Now, as
So,
Bonus
As
As
We think of
# = (-3x)/(x-absx sqrt(1+3/x))# .
For
# = lim_(xrarr-oo)(-3x)/(x+xsqrt(1+3/x))#
# = lim_(xrarr-oo)(-3)/(1+sqrt(1+3/x))#
# = -3/2# .
Explanation:
#sqrt(x^2+3x) >= 0# for all#x >= 0#
So
#lim_(x->+oo) (x+sqrt(x^2+3x)) >= lim_(x->+oo) x = +oo#
Much more interesting is the case
First notice that
To see this: Given
#(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2#
So if
#(t-epsilon)^2 = t^2 - 2tepsilon + epsilon^2 < t^2 - (C +epsilon^2) + epsilon^2 = t^2 - C#
So
#t-epsilon < sqrt(t^2-C) < sqrt(t^2) = t#
Let
Then:
#0 = lim_(t->oo) (sqrt(t^2-C) - t)#
#= lim_(x->-oo) (sqrt((x+3/2)^2 - 9/4) + x + 3/2)#
#= lim_(x->-oo) (sqrt(x^2+3x+9/4 - 9/4) + x + 3/2)#
#= lim_(x->-oo) (sqrt(x^2+3x) + x + 3/2)#
#= lim_(x->-oo) (sqrt(x^2+3x) + x) + 3/2#
Hence:
#lim_(x->-oo) (sqrt(x^2+3x) + x) = -3/2#