Prove the following
Question:

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}$

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

  2. does not exist

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

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


Correct Option: 1

Solution:

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}$

$=\lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^{4}}-2}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)}$

$=\lim _{y \rightarrow 0} \frac{\left(\sqrt{1+y^{4}}-1\right)\left(\sqrt{1+y^{4}}+1\right)}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}$

$=\lim _{y \rightarrow 0} \frac{1+y^{4}-1}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}$

$=\lim _{y \rightarrow 0} \frac{1}{\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}=\frac{1}{4 \sqrt{2}}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.