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

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(A) 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 the form is $1^

Vedclass · Accepted Answer

$e^4$

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(A) 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 the form is $1^\infty$.Here,$\mathop {\lim }\limits_{x 	o \infty } \frac{x^2 + 5x + 3}{x^2 + x + 3} = 1$.So,the expression is of the form $1^\infty$.Let $f(x) = \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 = \frac{x^2 + 5x + 3 - x^2 - x - 3}{x^2 + x + 3} = \frac{4x}{x^2 + x + 3}$.Then,$\mathop {\lim }\limits_{x 	o \infty } f(x)g(x) = \mathop {\lim }\limits_{x 	o \infty } \left( \frac{4x}{x^2 + x + 3} \cdot x ight) = \mathop {\lim }\limits_{x 	o \infty } \frac{4x^2}{x^2 + x + 3}$.Dividing numerator and denominator by $x^2$,we get $\mathop {\lim }\limits_{x 	o \infty } \frac{4}{1 + \frac{1}{x} + \frac{3}{x^2}} = \frac{4}{1 + 0 + 0} = 4$.Therefore,the limit is $e^4$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{{x^2} + 5x + 3}}{{{x^2} + x + 3}}} \right)^x} = $

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