Question

Question.
Consider the following.
f(x) = x − sqrtx((x − 1))
Estimate the limit of
f(x)as x approaches infinity and as x approaches negative infinity. Need help please:)

Answer 1

\[\lim_{x \rightarrow \infty}(x-\sqrt{x(x-1)}) =\lim_{x \rightarrow \infty}(x-\sqrt{x(x-1)}) \cdot \frac{x+\sqrt{x(x-1)}}{x+\sqrt{x(x-1)}} \\ =\lim_{x \rightarrow \infty}\frac{x^2-x(x-1)}{x+\sqrt{x(x-1)}}=\lim_{x \rightarrow \infty}\frac{x^2-x^2+x}{x+\sqrt{x(x-1)}}=\lim_{x \rightarrow \infty}\frac{x}{x+\sqrt{x^2-x}}\]
Try dividing top and bottom by sqrt(x^2)
sqrt(x^2)=x since x->infty (so x is positive)
--
for the problem you would sqrt(x^2)=-x since x->-infty (so x is negative)

Answer 2

However since it says estimate I wonder if would rather you use a numerical approach instead of an algebraic one.

Answer 3

I think numerical

Answer 4

x-> negative infitinity

Answer 5

Well see if the x-sqrt(x(x-1)) 's outputs are getting to some number as you enter in like
x=-20, x=-100, x=-1000

Answer 6

\[-20-\sqrt{-20(-20-1)}=? \\ -100-\sqrt{-100(-100-1)}=? \\ -1000-\sqrt{-1000(-1000-1)}=?\]

Answer 7

These are my outputs
-10^6 = -2000000.5000
-10^4 = -20000.5000
-10^2=-200.4988
10^0 = 1
10^2 = .5013
10^4 = .5000
10^6 = .5000

Answer 8

I don't know if these would help.

Answer 9

Ok notice for x->-infty, we have the y's aren't approaching some number
But for x->infty, we have the y's are approaching some number (that number being 1/2 or.5)

Answer 10

you could say as x->-infty, y->-infty

Answer 11

Because the y-values are getting negative huge

Answer 12

ahh I see now:)

Answer 13

Also I would write your one post up there as
These are my outputs
Let f(x)=x-sqrt(x(x-1))
f(-10^6) = -2000000.5000
f(-10^4) = -20000.5000
f(-10^2)=-200.4988

f(10^2) = .5013
f(10^4) = .5000
f(10^6) = .5000