Key Concepts and Formulas

• Indeterminate Forms: Understanding when a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, which suggests further manipulation or application of specific rules.
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} g(x) = 0$ or $\pm \infty$, and the limit of the ratio of their derivatives exists, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$.
• Chain Rule for Differentiation: Used to differentiate composite functions. If $h(x) = f(g(x))$, then $h'(x) = f'(g(x)) \cdot g'(x)$.
• Algebraic Manipulation for Limits: Techniques like multiplying by the conjugate can be used to simplify expressions and resolve indeterminate forms.

Step-by-Step Solution

Step 1: Evaluate the limit by direct substitution. We are asked to find the limit: $L = \mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$ Substituting $y=0$ into the expression, we get: $\frac{\sqrt{1 + \sqrt{1 + 0^4}} - \sqrt{2}}{0^4} = \frac{\sqrt{1 + \sqrt{1}} - \sqrt{2}}{0} = \frac{\sqrt{1+1} - \sqrt{2}}{0} = \frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0}$ This is an indeterminate form of type $\frac{0}{0}$, which indicates that we can use L'Hôpital's Rule or algebraic manipulation.

Step 2: Apply L'Hôpital's Rule. Since we have the indeterminate form $\frac{0}{0}$, we can apply L'Hôpital's Rule. We need to find the derivatives of the numerator and the denominator with respect to $y$.

Let $f(y) = \sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2$ and $g(y) = y^4$.

First, find the derivative of the numerator, $f'(y)$: $f'(y) = \frac{d}{dy} \left( \sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 \right)$ Using the chain rule, the derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}} \frac{du}{dy}$. Let $u = 1 + \sqrt{1+y^4}$. Then $\frac{du}{dy} = \frac{d}{dy}(1 + (1+y^4)^{1/2})$. $\frac{du}{dy} = 0 + \frac{1}{2}(1+y^4)^{-1/2} \cdot \frac{d}{dy}(1+y^4)$ $\frac{du}{dy} = \frac{1}{2\sqrt{1+y^4}} \cdot (4y^3) = \frac{2y^3}{\sqrt{1+y^4}}$ Now, applying the chain rule to the outer square root: $\frac{d}{dy} \left( \sqrt {1 + \sqrt {1 + {y^4}} } \right) = \frac{1}{2\sqrt {1 + \sqrt {1 + {y^4}} }} \cdot \frac{du}{dy}$ $f'(y) = \frac{1}{2\sqrt {1 + \sqrt {1 + {y^4}} }} \cdot \frac{2y^3}{\sqrt{1+y^4}} = \frac{y^3}{\sqrt {1 + \sqrt {1 + {y^4}} } \sqrt{1+y^4}}$ The derivative of the denominator, $g'(y)$: $g'(y) = \frac{d}{dy}(y^4) = 4y^3$

Now, apply L'Hôpital's Rule: L = \mathop {\lim }\limits_{y \to 0} \frac{f'(y)}{g'(y)} = \mathop {\lim }\limits_{y \to 0} \frac{\frac{y^3}{\sqrt {1 + \sqrt {1 + {y^4}} } \sqrt{1+y^4}}}{4y^3}}

Step 3: Simplify the expression after applying L'Hôpital's Rule. L = \mathop {\lim }\limits_{y \to 0} \frac{y^3}{\sqrt {1 + \sqrt {1 + {y^4}} } \sqrt{1+y^4}} \cdot \frac{1}{4y^3}} Cancel out the $y^3$ terms: $L = \mathop {\lim }\limits_{y \to 0} \frac{1}{4 \sqrt {1 + \sqrt {1 + {y^4}} } \sqrt{1+y^4}}$

Step 4: Evaluate the limit of the simplified expression. Now, substitute $y=0$ into the simplified expression: L = \frac{1}{4 \sqrt {1 + \sqrt {1 + 0^4}} \sqrt{1+0^4}}} L = \frac{1}{4 \sqrt {1 + \sqrt {1}} \sqrt{1}}} $L = \frac{1}{4 \sqrt {1 + 1} \cdot 1}$ $L = \frac{1}{4 \sqrt {2}}$

Alternative Approach using Conjugate Multiplication:

Step 1: Identify the indeterminate form and apply conjugate multiplication. As found earlier, the limit is of the form $\frac{0}{0}$. We can multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2$. L = \mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}} \times \frac{\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2}{\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2} $L = \mathop {\lim }\limits_{y \to 0} \frac{(\sqrt {1 + \sqrt {1 + {y^4}} })^2 - (\sqrt 2)^2}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)}$ $L = \mathop {\lim }\limits_{y \to 0} \frac{(1 + \sqrt {1 + {y^4}}) - 2}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)}$ $L = \mathop {\lim }\limits_{y \to 0} \frac{\sqrt {1 + {y^4}} - 1}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)}$

Step 2: Apply conjugate multiplication again to the new numerator. The expression is still of the form $\frac{0}{0}$ if we substitute $y=0$ (since $\sqrt{1+0}-1 = 0$ and $0^4=0$). So, we multiply the numerator and denominator by the conjugate of $\sqrt{1+y^4}-1$, which is $\sqrt{1+y^4}+1$. $L = \mathop {\lim }\limits_{y \to 0} \frac{(\sqrt {1 + {y^4}} - 1)}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)} \times \frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1}$ $L = \mathop {\lim }\limits_{y \to 0} \frac{(\sqrt {1 + {y^4}})^2 - 1^2}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)(\sqrt{1+y^4}+1)}$ $L = \mathop {\lim }\limits_{y \to 0} \frac{(1 + y^4) - 1}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)(\sqrt{1+y^4}+1)}$ $L = \mathop {\lim }\limits_{y \to 0} \frac{y^4}{y^4 (\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)(\sqrt{1+y^4}+1)}$

Step 3: Simplify and evaluate the limit. Cancel out the $y^4$ terms: $L = \mathop {\lim }\limits_{y \to 0} \frac{1}{(\sqrt {1 + \sqrt {1 + {y^4}} } + \sqrt 2)(\sqrt{1+y^4}+1)}$ Now, substitute $y=0$: $L = \frac{1}{(\sqrt {1 + \sqrt {1 + 0^4}} + \sqrt 2)(\sqrt{1+0^4}+1)}$ $L = \frac{1}{(\sqrt {1 + \sqrt {1}} + \sqrt 2)(\sqrt{1}+1)}$ $L = \frac{1}{(\sqrt {1 + 1} + \sqrt 2)(1+1)}$ $L = \frac{1}{(\sqrt {2} + \sqrt 2)(2)}$ $L = \frac{1}{(2\sqrt {2})(2)}$ $L = \frac{1}{4\sqrt {2}}$

Common Mistakes & Tips

• Incorrect Application of L'Hôpital's Rule: Ensure the limit is indeed in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Also, differentiate the numerator and denominator separately, not as a quotient.
• Errors in Differentiation: The chain rule can be tricky. Carefully differentiate nested functions, paying attention to each layer of the composite function. For example, differentiating $\sqrt{1+\sqrt{1+y^4}}$ requires multiple applications of the chain rule.
• Algebraic Simplification Errors: When using conjugate multiplication, ensure all terms are expanded and simplified correctly. Canceling terms prematurely or incorrectly can lead to the wrong answer.

Summary

The given limit results in an indeterminate form of $\frac{0}{0}$ upon direct substitution. This can be resolved using either L'Hôpital's Rule or algebraic manipulation involving multiplying by conjugates. Both methods require careful application of differentiation rules (specifically the chain rule for L'Hôpital's) or algebraic simplification. After applying these techniques and simplifying the expression, we evaluate the limit by substituting $y=0$, yielding the result $\frac{1}{4\sqrt{2}}$.

The final answer is $\boxed{\frac{1}{4\sqrt 2}}$.