_{y arrow 0} {{1+{1+y^{4}}}-{2}}{y^{4}}=

Correct Answer is : 142{1}{4 {2}}421

Correct Answer is : 142{1}{4 {2}}421

Test Index

MHTCET 16 May 2024 Shift 2 Paper

Question : 116 of 150

Marks: +1, -0

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

Solution:

Concept:

Rationalise the numerator twice to remove the radicals, then evaluate the limit by direct substitution.

Explanation:

Let

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

Rationalise the numerator with the conjugate

\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}

L = \lim_{y\to0} \frac{\left(1+\sqrt{1+y^{4}}\right)-2}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)}

= \lim_{y\to0} \frac{\sqrt{1+y^{4}}-1}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)}

Now rationalise

\sqrt{1+y^{4}}-1

with its conjugate

\sqrt{1+y^{4}}+1

= \lim_{y\to0} \frac{y^{4}}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}

Cancel

y^{4}

and substitute

y=0

\frac{1}{\left(\sqrt{1+\sqrt{1+0}}+\sqrt{2}\right)\left(\sqrt{1+0}+1\right)}

= \frac{1}{\left(\sqrt{2}+\sqrt{2}\right)(2)} = \frac{1}{4\sqrt{2}}

Answer:

\frac{1}{4\sqrt{2}}

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