First slide

Introduction to limits

Question

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

Moderate

A

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

does not exist
C

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

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

Solution

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

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