mathop {lim }limits_{x to - infty } frac{{sqr | vedclass

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(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$$x 	o -\infty$ હોવાથી,આપણે $x = -|x| = -\sqrt{x^

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$-1/2$

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(A) આપેલ લક્ષ: $\mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$$x 	o -\infty$ હોવાથી,આપણે $x = -|x| = -\sqrt{x^2}$ લખી શકીએ.અંશ અને છેદને $|x|$ વડે ભાગતા (જ્યાં $x $= \mathop {\lim }\limits_{x 	o - \infty } \frac{{\frac{{\sqrt {4{x^2} + 5x + 8} }}{{|x|}}}}{{\frac{{4x + 5}}{{|x|}}}} = \mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4 + \frac{5}{x} + \frac{8}{{{x^2}}}} }}{{ - \left( {4 + \frac{5}{x}} ight)}}$જેમ $x 	o -\infty$,તેમ $\frac{1}{x} 	o 0$ અને $\frac{1}{{{x^2}}} 	o 0$.$= \frac{{\sqrt {4 + 0 + 0} }}{{ - (4 + 0)}} = \frac{2}{{ - 4}} = - \frac{1}{2}$.

$\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ ની કિંમત શોધો.

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