If mathop {lim }limits_{x to infty } {left( { | vedclass

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(B) We know that $\mathop {\lim }\limits_{x 	o \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o \infty } f(x)g(x)}$ if $\mathop {\lim

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$a = 1, b \in \mathbb{R}$

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(B) We know that $\mathop {\lim }\limits_{x 	o \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o \infty } f(x)g(x)}$ if $\mathop {\lim }\limits_{x 	o \infty } f(x) = 0$.Given $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} ight)^{2x}} = {e^2}$.Applying the formula,we get $e^{\mathop {\lim }\limits_{x 	o \infty } (\frac{a}{x} + \frac{b}{{{x^2}}}) \cdot 2x} = {e^2}$.This simplifies to $e^{\mathop {\lim }\limits_{x 	o \infty } (2a + \frac{2b}{x})} = {e^2}$.Evaluating the limit,we get $e^{2a + 0} = {e^2}$.Thus,$e^{2a} = {e^2}$,which implies $2a = 2$,so $a = 1$.Since the term $\frac{b}{x^2}$ vanishes as $x 	o \infty$ regardless of the value of $b$,$b$ can be any real number.Therefore,$a = 1$ and $b \in \mathbb{R}$.

If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} + \frac{b}{{{x^2}}}} \right)^{2x}} = {e^2},$ then the values of $a$ and $b$ are

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