The value of $$\lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4}$$ is:

We have to evaluate the limit

$$L=\lim_{y\to 0}\dfrac{\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt2}{y^{4}}.$$

Because the denominator contains the very small power $$y^{4}$$, it is natural to expand every radical in the numerator by using the binomial (Taylor) series of the square-root function near $$1$$.

First, recall the standard expansion valid for $$|t|\ll 1$$:

$$\sqrt{1+t}=1+\dfrac{t}{2}-\dfrac{t^{2}}{8}+O(t^{3}).$$

We apply this formula step by step. Put $$t=y^{4}$$ in it. Then

$$\sqrt{\,1+y^{4}\,}=1+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12}).$$

Now add $$1$$ to the above expression because it appears inside the outer radical:

$$1+\sqrt{\,1+y^{4}\,}=1+\left(1+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12})\right) =2+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}+O(y^{12}).$$

The next task is to take the square root of this new quantity. For convenience we factor out the constant $$2$$:

$$2+\dfrac{y^{4}}{2}-\dfrac{y^{8}}{8}=2\Bigl(1+\underbrace{\dfrac{y^{4}}{4}-\dfrac{y^{8}}{16}+O(y^{12})}_{\displaystyle v}\Bigr).$$

Thus

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,} =\sqrt{2}\;\sqrt{1+v}, \qquad v=\dfrac{y^{4}}{4}-\dfrac{y^{8}}{16}+O(y^{12}).$$

We again employ the same series $$\sqrt{1+v}=1+\dfrac{v}{2}-\dfrac{v^{2}}{8}+O(v^{3})$$. Substituting the expression of $$v$$ and keeping only terms up to order $$y^{4}$$ (higher powers will vanish when divided by $$y^{4}$$ in the limit) we get

$$\sqrt{1+v}=1+\dfrac{1}{2}\left(\dfrac{y^{4}}{4}\right)+O(y^{8}) =1+\dfrac{y^{4}}{8}+O(y^{8}).$$

Therefore

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,} =\sqrt{2}\Bigl(1+\dfrac{y^{4}}{8}+O(y^{8})\Bigr) =\sqrt{2}+\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8}).$$

Now form the needed numerator:

$$\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt{2} =\left(\sqrt{2}+\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8})\right)-\sqrt{2} =\dfrac{\sqrt{2}\,y^{4}}{8}+O(y^{8}).$$

Divide this by $$y^{4}$$:

$$\dfrac{\sqrt{\,1+\sqrt{\,1+y^{4}\,}\,}-\sqrt{2}}{y^{4}} =\dfrac{\sqrt{2}}{8}+O(y^{4}).$$

As $$y\to 0$$ the term $$O(y^{4})$$ tends to $$0$$, hence the limit equals the constant coefficient just obtained:

$$L=\dfrac{\sqrt{2}}{8}=\dfrac{1}{4\sqrt{2}}.$$

Hence, the correct answer is Option B.