Use calculus to evaluate (lhopitals rule) lim x>infinity x^2 - ln(2/x)/3x^2 + 2x

asked Jun 16, 2022 98.9k views

20 votes

Use calculus to evaluate (lhopitals rule)
lim x>infinity x^2 - ln(2/x)/3x^2 + 2x

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4 votes

Answer:

1/3

Explanation:

$[x^2 - ln(2/x)]/[3x^2 + 2x]\\$

∞/∞

$(d)/(dx)[x^2-ln(2/x)] =2x+(1)/(x)\\ (d)/(dx)[3x^2+2x]=6x+2\\(2x+(1)/(x))/(6x+2)$

Also ∞/∞

$(d)/(dx)[2x+(1)/(x)]=2-1/x^2\\(d)/(dx)[6x+2]=6\\\\$

$\lim_(x \to \infty) (2-(1)/(x^2) )/(6)\\=(2)/(6) \\=(1)/(3)$

Nodrog answered Jun 20, 2022

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