Mathematics - Limits Question with Solution | TestHub

To solve this question, we need to apply the concept of rationalization twice.

Given limit is$\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} ( \frac{0}{0} \mathrm{form})$

$=\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}\times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$

$=\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{\left(\sqrt{1+y^4}-1\right)}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)}\times \frac{\left(\sqrt{1+y^4}+1\right)}{\left(\sqrt{1+y^4}+1\right)}$

$=\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{y^4}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)}$

$=\underset{y\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{1}{\left(\sqrt{1+\sqrt{1+y^4} }+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)}$

$=\frac{1}{4\sqrt{2}}$.