A) exists and equals \[\frac{1}{2\sqrt{2}}\]
B) exists and equals \[\frac{1}{4\sqrt{2}}\]
C) exists and equals \[\frac{1}{2\sqrt{2}(\sqrt{2}+1)}\]
D) does not exist
Correct Answer: B
Solution :
\[\underset{y\to 0}{\mathop{\lim }}\,\,\,\frac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}}\] Rationalise \[=\underset{y\to 0}{\mathop{\,\,\,\,\lim }}\,\,\,\frac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{{{y}^{4}}}\,\,\times \,\,\frac{\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}}{\sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2}}\] \[=\underset{y\to 01}{\mathop{\,\,\,\,\lim }}\,\,\,\frac{\sqrt{1+\sqrt{1+{{y}^{4}}}}-\sqrt{2}}{\left( \sqrt{1+\sqrt{1+{{y}^{4}}}}+\sqrt{2} \right){{y}^{4}}}\,\,\] \[=\,\,\,\underset{y\,\to \,0}{\mathop{\,\,\,\,\lim }}\,\,\,\frac{\sqrt{1+{{y}^{4}}}\,-1}{2\sqrt{2}\,{{y}^{4}}}\,\,\] \[=\,\,\,\frac{1}{2\sqrt{2}}\underset{y\,\to \,0}{\mathop{\,\,\,\lim }}\,\,\,\frac{\left( \sqrt{1+{{y}^{4}}}\,-1 \right)\left( \sqrt{1+{{y}^{4}}}+1 \right)}{\left( \sqrt{1+{{y}^{4}}}+1 \right)\,.\,{{y}^{4}}}\,\,\] \[=\,\,\,\frac{1}{2\sqrt{2}}\underset{y\,\to \,0}{\mathop{\,\,\lim }}\,\,\,\frac{1+{{y}^{4}}-1}{{{y}^{4}}\,\,\left( \sqrt{1+{{y}^{4}}}+1 \right)}\,\,\] \[=\,\,\,\,\frac{1}{2\sqrt{2}}\,\,\times \,\,\frac{1}{2}\,\,=\,\,\frac{1}{4\sqrt{2}}\]
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