The value of mathop {lim }limits_{x to - inft | vedclass

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Question

(A) Given the limit: $\mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$Since $x 	o -\infty$,we can write $x = -|x

Vedclass · Accepted Answer

$-1/2$

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(A) Given the limit: $\mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$Since $x 	o -\infty$,we can write $x = -|x| = -\sqrt{x^2}$.Dividing the numerator and denominator by $|x|$ (where $|x| = -x$ for $x $= \mathop {\lim }\limits_{x 	o - \infty } \frac{{\frac{{\sqrt {4{x^2} + 5x + 8} }}{{|x|}}}}{{\frac{{4x + 5}}{{|x|}}}} = \mathop {\lim }\limits_{x 	o - \infty } \frac{{\sqrt {4 + \frac{5}{x} + \frac{8}{{{x^2}}}} }}{{ - \left( {4 + \frac{5}{x}} ight)}}$As $x 	o -\infty$,$\frac{1}{x} 	o 0$ and $\frac{1}{{{x^2}}} 	o 0$.$= \frac{{\sqrt {4 + 0 + 0} }}{{ - (4 + 0)}} = \frac{2}{{ - 4}} = - \frac{1}{2}$.

The value of $\mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {4{x^2} + 5x + 8} }}{{4x + 5}}$ is

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