@Background Pony #ED9E
The limits exist at places other than +∞. They’re just not terribly interesting. For any finite value, you simply evaluate the expression at that point. And on the negative side, the value grows without bound as you approach -∞.
@TPC132 I have a mighty need! (No, seriously, I do. There’s a whole spiel on googology I want to post in the Stupid Stuff thread, but I can’t without proper notation.)
The expression √(4n² + 4n + 1) simplifies to (2n + 1) for positive n, so the denominator simplifies to 1.
For the numerator: the +1 in √(9n² + 1) can be omitted in the limit because the difference between √(x + 1) and √x is negligible. The difference can be estimated as:
√(x + 1) - √x = ((√(x + 1) - √x)(√(x + 1) + √x)) / (√(x + 1) + √x) = (x + 1 - x) / (√(x + 1) + √x) < 1 / (2√x)
This becomes arbitrarily small for large x. Visually, it can also be understood as a small increment to the √x function plot when x is increased by 1, limited by the decreasing slope of the tangent line (derivative 1/(2√x) approaches zero for large x).
Thus, √(9n² + 1) for large n can be simplified to √(9n²). Therefore, the numerator can be simplified to:
√(9n²) - 3n - 3 = -3
The entire fraction can then be simplified to:
-3/1 = -3
This “simplification” could be performed more formally using the standard properties of limits, where the limit of a quotient is the quotient of the limits. The same applies for addition and subtraction. However, √(x + 1) requires a bit of additional work (or L’Hôpital’s Rule).
Additionally, the notation n → ∞ should be placed under the limit, otherwise, the problem is stated incorrectly.
The limits exist at places other than +∞. They’re just not terribly interesting. For any finite value, you simply evaluate the expression at that point. And on the negative side, the value grows without bound as you approach -∞.
I have a mighty need! (No, seriously, I do. There’s a whole spiel on googology I want to post in the Stupid Stuff thread, but I can’t without proper notation.)
Derpibooru LaTeX integration when
√(x + 1) - √x = ((√(x + 1) - √x)(√(x + 1) + √x)) / (√(x + 1) + √x) = (x + 1 - x) / (√(x + 1) + √x) < 1 / (2√x)
This becomes arbitrarily small for large x. Visually, it can also be understood as a small increment to the √x function plot when x is increased by 1, limited by the decreasing slope of the tangent line (derivative 1/(2√x) approaches zero for large x).
√(9n²) - 3n - 3 = -3
-3/1 = -3
Edited