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Also how is doubling time calculated in general. Show previous comments 7 more

let me try...

1/x < 2
we can see that 1/x = 2 when x = 1/2, and that increasing x will decrease 1/x, so the answer is
x > 1/2 I guess
or maybe you can just multiply both parts by x and then divide by 2, but dunno if you're allowed to do that??? maybe it will get tricky since x might be either positive or negative, dunno

classic method here, multiply numerator and denominator by something so you can use the formula (xy)*(x+y)=x^2  y^2
lim x→∞ (x − √ (x^2 + x − 1)) = lim x→∞ ((x^2  x^2  x + 1) / (x + √ (x^2 + x  1)) = lim x→∞ ((1  x) / (x + √ (x^2 + x − 1))) = lim x→∞ ((1 + 1/x) / (1 + √ (1 + 1/x  1/x^2))) = 1 / (1 + 1) = 1/2.

x^3  x  1 = 0
no idea, either I'm missing something trivial or you just have to take a lucky guess? 
Memfis said:
let me try...

1/x < 2
we can see that 1/x = 2 when x = 1/2, and that increasing x will decrease 1/x, so the answer is
x > 1/2 I guess
or maybe you can just multiply both parts by x and then divide by 2, but dunno if you're allowed to do that??? maybe it will get tricky since x might be either positive or negative, dunno

I actually solved this before (x < 0 or x > 1/2), but I wanted to double check.
x^3  x  1 = 0
no idea, either I'm missing something trivial or you just have to take a lucky guess?
There is no "regular" way to solve this, instead there is the cubic general formula.