limy→0√(1+√(1+y4)) - √2y4

Let the given limit be L.
L = limy→0 √(1+√(1+y4)) - √2y4
Let f(y) = √(1+√(1+y4))
f(0) = √(1+√1) = √2
Let g(y) = √2y4
g(0) = 0
Then L = limy→0 f(y) - g(y) = f(0) - g(0) = √2 - 0 = √2
However, this is incorrect. Let's use L'Hopital's rule.
Let's consider the expression:
√(1+√(1+y4)) - √2y4
Let's use the binomial approximation (1+x)n ≈ 1+nx for small x.
√(1+y4) ≈ 1 + y4/2
√(1+√(1+y4)) ≈ √(1 + 1 + y4/2) = √(2 + y4/2) ≈ √2(1 + y4/4)1/2 ≈ √2(1 + y4/8)
So, √(1+√(1+y4)) - √2y4 ≈ √2(1 + y4/8) - √2y4 = √2(1 + y4/8 - y4) = √2(1 - 7y4/8)
As y→0, the expression tends to √2.
However, the limit is not √2. Let's rationalize the expression.
Let x = y4.
Then we have limx→0 √(1+√(1+x)) - √2x = limx→0 [√(1+√(1+x)) - √2x] * [√(1+√(1+x)) + √2x] / [√(1+√(1+x)) + √2x]
= limx→0 (1+√(1+x) - 2x) / [√(1+√(1+x)) + √2x]
Using binomial expansion,
1+√(1+x) ≈ 1 + 1 + x/2 = 2 + x/2
So, the numerator becomes 2 + x/2 - 2x = 2 - 3x/2
The denominator tends to 2√2 as x→0.
Thus, the limit is (2 - 0)/(2√2) = 1/√2 = √2/2 = 1/2√2
Therefore, the limit exists and equals 1/2√2.