\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}=\] [IIT 1996; DCE 2001]

\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{

1.	\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}=\] [IIT 1996; DCE 2001]
A.	\[{{e}^{2}}\]
B.	\[e\]
C.	\[{{e}^{-2}}\]
D.	\[{{e}^{-1}}\]
Answer» B. \[e\]

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\[\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}=\] [IIT 1996; DCE 2001]

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