\mathop \lim \limits_y \to 0 \sqrt 1 + \sqrt 1 + y^4 - \sqrt 2 \over y^4

exists and equals 1 \over 4\sqrt 2

exists and equals 1 \over 2\sqrt 2

exists and equals 1 \over 2\sqrt 2 (1 + \sqrt 2)

Limits Continuity and Differentiability PYQ — JEE Main 2019 · 9 Jan · Shift 1 · Q34 with Solution

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Correct answer: B

We need to evaluate $L=\lim_{y\to 0}\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt2}{y^4}.$
Since the expression is of the form $\frac{0}{0}$ , let us simplify it using standard limit expansion.

First, use $\sqrt{1+t}=1+\frac t2+o(t) \quad (t\to 0).$

Here $t=y^4$ , so $\sqrt{1+y^4}=1+\frac{y^4}{2}+o(y^4).$

Therefore, $1+\sqrt{1+y^4}=2+\frac{y^4}{2}+o(y^4).$

Now expand the outer square root around $2$ : $\sqrt{2+h}=\sqrt2+\frac{h}{2\sqrt2}+o(h) \quad (h\to 0).$

Here $h=\frac{y^4}{2}+o(y^4).$

So,

=\sqrt2+\frac{1}{2\sqrt2}\left(\frac{y^4}{2}+o(y^4)\right)+o(y^4).$$ Thus, $$\sqrt{1+\sqrt{1+y^4}}-\sqrt2 =\frac{y^4}{4\sqrt2}+o(y^4).$$ 4. Divide by $y^4$: $$\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt2}{y^4} =\frac{1}{4\sqrt2}+o(1).$$ Hence, $$L=\frac{1}{4\sqrt2}.$$ 5. Option check: - A: $\frac{1}{2\sqrt2}$ — incorrect - B: $\frac{1}{4\sqrt2}$ — correct - C: $\frac{1}{2\sqrt2(1+\sqrt2)}$ — incorrect - D: does not exist — incorrect Therefore, the limit exists and equals $$\boxed{\frac{1}{4\sqrt2}}.$$