`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 can follow these steps:

### Step 1: Substitute \( t = \sqrt{1 + y^4} \)

Let \( t = \sqrt{1 + y^4} \). Then, as \( y \to 0 \), \( t \to \sqrt{1 + 0} = 1 \).

### Step 2: Express \( y^4 \) in terms of \( t \)

From \( t = \sqrt{1 + y^4} \), we can square both sides to get:
\[ t^2 = 1 + y^4 \implies y^4 = t^2 - 1 \]

### Step 3: Rewrite the limit in terms of \( t \)

Now we can rewrite the limit:
\[ \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4} = \lim_{t \to 1} \frac{\sqrt{1 + t} - \sqrt{2}}{t^2 - 1} \]

### Step 4: Evaluate the limit

Substituting \( t = 1 \) gives us an indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and the denominator.

#### Differentiate the numerator:
\[ \frac{d}{dt}(\sqrt{1 + t}) = \frac{1}{2\sqrt{1 + t}} \]

#### Differentiate the denominator:
\[ \frac{d}{dt}(t^2 - 1) = 2t \]

### Step 5: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:
\[ \lim_{t \to 1} \frac{\frac{1}{2\sqrt{1 + t}}}{2t} = \lim_{t \to 1} \frac{1}{4t\sqrt{1 + t}} \]

### Step 6: Substitute \( t = 1 \)

Substituting \( t = 1 \):
\[ \frac{1}{4 \cdot 1 \cdot \sqrt{1 + 1}} = \frac{1}{4 \cdot 1 \cdot \sqrt{2}} = \frac{1}{4\sqrt{2}} \]

### Final Answer

Thus, the limit is:
\[ \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4} = \frac{1}{4\sqrt{2}} \]