investigate the limit numerically and graphically.\n lim_{x\rightarrowpminfty}\frac{6x + 1}{sqrt{4x^{2}+9}}…

investigate the limit numerically and graphically.\n lim_{x\rightarrowpminfty}\frac{6x + 1}{sqrt{4x^{2}+9}} \ncalculate the values of ( f(x)=\frac{6x + 1}{sqrt{4x^{2}+9}} ) for ( x=pm100,pm500,pm1000 ), and ( pm10000 ).\n(use decimal notation. give your answers to six decimal places.)\n\n( f(-100)=)\n\n( f(-500)=)

investigate the limit numerically and graphically.\n lim_{x\rightarrowpminfty}\frac{6x + 1}{sqrt{4x^{2}+9}} \ncalculate the values of ( f(x)=\frac{6x + 1}{sqrt{4x^{2}+9}} ) for ( x=pm100,pm500,pm1000 ), and ( pm10000 ).\n(use decimal notation. give your answers to six decimal places.)\n\n( f(-100)=)\n\n( f(-500)=)

Answer

Explanation:

Step1: Substitute x = - 100

Substitute $x=-100$ into $f(x)=\frac{6x + 1}{\sqrt{4x^{2}+9}}$. $f(-100)=\frac{6\times(-100)+1}{\sqrt{4\times(-100)^{2}+9}}=\frac{-600 + 1}{\sqrt{40000+9}}=\frac{-599}{\sqrt{40009}}\approx\frac{-599}{200.0225}\approx - 2.994657$

Step2: Substitute x = - 500

Substitute $x = - 500$ into $f(x)=\frac{6x + 1}{\sqrt{4x^{2}+9}}$. $f(-500)=\frac{6\times(-500)+1}{\sqrt{4\times(-500)^{2}+9}}=\frac{-3000+1}{\sqrt{1000000 + 9}}=\frac{-2999}{\sqrt{1000009}}\approx\frac{-2999}{1000.0045}\approx - 2.998991$

Answer:

$f(-100)\approx - 2.994657$ $f(-500)\approx - 2.998991$