limy→0 1+1+y4−2y4

# $\underset{y\to 0}{lim} \frac{\sqrt{1+\sqrt{1+{y}^{4}}}-\sqrt{2}}{{y}^{4}}$

1. A

exists and equals $\frac{1}{4\sqrt{2}}$

2. B

does not exist

3. C

exists and equals $\frac{1}{2\sqrt{2}}$

4. D

exists and equals $\frac{1}{2\sqrt{2\left(\sqrt{2}+1\right)}}$

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### Solution:

$\underset{y\to 0}{lim} \frac{\sqrt{1+\sqrt{1+{y}^{4}}}-\sqrt{2}}{{y}^{4}}$

$\begin{array}{l}=\underset{y\to 0}{lim} \frac{1+\sqrt{1+{y}^{4}}-2}{{y}^{4}\left(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}\right)}\\ =\underset{y\to 0}{lim} \frac{\left(\sqrt{1+{y}^{4}}-1\right)\left(\sqrt{1+{y}^{4}}+1\right)}{{y}^{4}\left(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}\right)\left(\sqrt{1+{y}^{4}}+1\right)}\end{array}$

$\begin{array}{l}=\underset{y\to 0}{lim} \frac{1+{y}^{4}-1}{{y}^{4}\left(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}\right)\left(\sqrt{1+{y}^{4}}+1\right)}\\ =\underset{y\to 0}{lim} \frac{1}{\left(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}\right)\left(\sqrt{1+{y}^{4}}+1\right)}=\frac{1}{4\sqrt{2}}\end{array}$

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