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

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Question

(A) આપણે જાણીએ છીએ કે જો સ્વરૂપ $1^\infty$ હોય,તો $\mathop {\lim }\limits_{x 	o \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o \inft

Vedclass · Accepted Answer

$e^4$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે જો સ્વરૂપ $1^\infty$ હોય,તો $\mathop {\lim }\limits_{x 	o \infty } {(1 + f(x))^{g(x)}} = e^{\mathop {\lim }\limits_{x 	o \infty } f(x)g(x)}$ થાય.અહીં,$\mathop {\lim }\limits_{x 	o \infty } \frac{x^2 + 5x + 3}{x^2 + x + 3} = 1$ છે.તેથી,આ પદાવલિ $1^\infty$ સ્વરૂપમાં છે.ધારો કે $f(x) = \frac{x^2 + 5x + 3}{x^2 + x + 3} - 1 = \frac{4x}{x^2 + x + 3}$.ત્યારબાદ,$\mathop {\lim }\limits_{x 	o \infty } f(x)g(x) = \mathop {\lim }\limits_{x 	o \infty } \left( \frac{4x}{x^2 + x + 3} \cdot x ight) = \mathop {\lim }\limits_{x 	o \infty } \frac{4x^2}{x^2 + x + 3}$.અંશ અને છેદને $x^2$ વડે ભાગતા,આપણને $\mathop {\lim }\limits_{x 	o \infty } \frac{4}{1 + \frac{1}{x} + \frac{3}{x^2}} = \frac{4}{1 + 0 + 0} = 4$ મળે છે.તેથી,લક્ષનું મૂલ્ય $e^4$ છે.

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

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