JEE Main 2019 — Mathematics Calculus

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

Given limit is $\underset{y\rightarrow 0}{lim}\frac{\sqrt{1+\sqrt{1+{y}^{4}}}-\sqrt{2}}{{y}^{4}}(\frac{0}{0}\mathrm{form})$y→0lim​y41+1+y4​​−2​​(00​form)

$$=\underset{y\rightarrow 0}{lim}\frac{\sqrt{1+\sqrt{1+{y}^{4}}}-\sqrt{2}}{{y}^{4}}\times \frac{\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}}{\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2}}$$=y→0lim​y41+1+y4​​−2​​×1+1+y4​​+2​1+1+y4​​+2​​

$$=\underset{y\rightarrow 0}{lim}\frac{(\sqrt{1+{y}^{4}}-1)}{{y}^{4}(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2})}\times \frac{(\sqrt{1+{y}^{4}}+1)}{(\sqrt{1+{y}^{4}}+1)}$$=y→0lim​y4(1+1+y4​​+2​)(1+y4​−1)​×(1+y4​+1)(1+y4​+1)​

$$=\underset{y\rightarrow 0}{lim}\frac{{y}^{4}}{{y}^{4}(\sqrt{1+\sqrt{1+{y}^{4}}}+\sqrt{2})(\sqrt{1+{y}^{4}}+1)}$$=y→0lim​y4(1+1+y4​​+2​)(1+y4​+1)y4​

$$=\underset{y\rightarrow 0}{lim}\frac{1}{(\sqrt{1+\sqrt{1+{y}^{4}} }+\sqrt{2})(\sqrt{1+{y}^{4}}+1)}$$=y→0lim​(1+1+y4​​+2​)(1+y4​+1)1​

$=\frac{1}{4\sqrt{2}}$=42​1​.