Bantu 2 itu yaa yg bisa:’)

By Admin ReiPosted on December 5, 2022

Bantu 2 itu yaa yg bisa:')

Bantu 2 itu yaa yg bisa:’)

$green{huge{9.}}$

$lim_{xto infty}~frac{3x^4-5x^2+4}{6x^4+7x^3-8}$

$=lim_{xto infty}~frac{frac{3x^4-5x^2+4}{x^4}}{frac{6x^4+7x^3-8}{x^4}}$

$=lim_{xto infty}~frac{3-frac{5}{x^2}+frac{4}{x^4}}{6+frac{7}{x}-frac{8}{x^4}}$

$=frac{3-frac{5}{infty^2}+frac{4}{infty^4}}{6+frac{7}{infty}-frac{8}{infty^4}}$

$=frac{3-0+0}{6+0-0}=frac{3}{6}=pink{huge{frac{1}{2}}}$

$huge{to sf (~red{A}~)}$

$\text{[math]}$

$green{huge{10.}}$

$lim_{xto infty}~left(sqrt{4x^2-8x+3}-(2x-1)right)$

$=lim_{xto infty}~left(sqrt{4x^2-8x+3}-(2x-1)times frac{sqrt{4x^2-8x+3}+(2x-1)}{sqrt{4x^2-8x+3}+(2x-1)}right)$

$=lim_{xto infty}~left(frac{left(sqrt{4x^2-8x+3}right)^2-(2x-1)^2}{sqrt{4x^2-8x+3}+(2x-1)}right)$

$=lim_{xto infty}~left(frac{left(4x^2-8x+3right)-left(4x^2-4x+1right)}{sqrt{4x^2-8x+3}+(2x-1)}right)$

$=lim_{xto infty}~left(frac{4x^2-8x+3-4x^2+4x-1}{sqrt{4x^2-8x+3}+(2x-1)}right)$

$=lim_{xto infty}~left(frac{-4x+2}{sqrt{4x^2-8x+3}+(2x-1)}right)$

$=lim_{xto infty}~left(frac{frac{-4x+2}{x}}{frac{sqrt{4x^2-8x+3}+(2x-1)}{x}}right)$

$=lim_{xto infty}~left(frac{frac{-4x+2}{x}}{sqrt{frac{4x^2-8x+3}{x^2}}+left(frac{2x-1}{x}right)}right)$

$=lim_{xto infty}~left(frac{-4+frac{2}{x}}{sqrt{4-frac{8}{x}+frac{3}{x^2}}+left(2-frac{1}{x}right)}right)$

$=frac{-4+frac{2}{infty}}{sqrt{4-frac{8}{infty}+frac{3}{infty^2}}+left(2-frac{1}{infty}right)}$

$=frac{-4+0}{sqrt{4-0+0}+left(2-0right)}=frac{-4}{sqrt{4}+2}=frac{-4}{2+2}$

$=frac{-4}{4}=pink{huge{-1}}$

$huge{to sf (~red{B}~)}$