`lim_(yto0) (sqrt(1+sqrt(1+y^4))-sqrt2)/y^4`

Options

A. exists and equals `1/(4sqrt2)`
B. does not exist
C. exists and equals `1/(2sqrt2)`
D. exists and equals `1/(2sqrt2(sqrt2+1))`

Answer

To solve the limit

\[ \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4}, \]

we will follow these steps:

### Step 1: Multiply by the Conjugate
To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator:

\[ \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4} \cdot \frac{\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2}}. \]

This gives us:

\[ \frac{(1 + \sqrt{1 + y^4}) - 2}{y^4(\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2})}. \]

### Step 2: Simplify the Numerator
The numerator simplifies to:

\[ 1 + \sqrt{1 + y^4} - 2 = \sqrt{1 + y^4} - 1. \]

So we rewrite the limit as:

\[ \lim_{y \to 0} \frac{\sqrt{1 + y^4} - 1}{y^4(\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2})}. \]

### Step 3: Use the Binomial Expansion
For small values of \(y\), we can use the binomial expansion for \(\sqrt{1 + y^4}\):

\[ \sqrt{1 + y^4} \approx 1 + \frac{y^4}{2}. \]

Thus,

\[ \sqrt{1 + y^4} - 1 \approx \frac{y^4}{2}. \]

### Step 4: Substitute Back into the Limit
Substituting this back into our limit gives:

\[ \lim_{y \to 0} \frac{\frac{y^4}{2}}{y^4(\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2})}. \]

### Step 5: Cancel \(y^4\)
The \(y^4\) terms cancel out:

\[ \lim_{y \to 0} \frac{1/2}{\sqrt{1 + \sqrt{1 + y^4}} + \sqrt{2}}. \]

### Step 6: Evaluate the Limit
As \(y \to 0\), \(\sqrt{1 + \sqrt{1 + y^4}} \to \sqrt{1 + \sqrt{1}} = \sqrt{2}\). Therefore, the limit simplifies to:

\[ \frac{1/2}{\sqrt{2} + \sqrt{2}} = \frac{1/2}{2\sqrt{2}} = \frac{1}{4\sqrt{2}}. \]

### Final Answer
Thus, the final answer is:

\[ \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4} = \frac{1}{4\sqrt{2}}. \]

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