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Answer : In your very first step, you cannot break a limit into the subtraction of two limits unless both of the other limits exist. The theorem: lim ⁡ x → a ( f ( x ) + g ( x ) ) = lim ⁡ x → a f ( x ) + lim ⁡ x → a g ( x ) \lim _{x\to a} (f(x)+g(x))=\lim _{x\to a}f(x) +\lim _{x\to a}g(x) lim x → a ​ ( f ( x ) + g ( x )) = lim x → a ​ f ( x ) + lim x → a ​ g ( x ) is ONLY valid if the two limits on the right hand side exist. In your case, the second limit clearly does not exist, because it goes to infinity. Edit for clarity, neither does the first limit. So in effect, what you tried to do was make this an ∞ − ∞ \infty - \infty ∞ − ∞ , which doesn't work as seperate limits, but does work together (sometimes)

Answer : Question 1: your way is not correct (reason: see my answer to question 2). Question 2: ∫ 0 ∞ \int_0^{\infty} ∫ 0 ∞ ​ is convergent    ⟺    \iff ⟺ the integrals ∫ 0 1 , ∫ 1 3 / 2 \int_0^{1},\int_{1}^{3/2} ∫ 0 1 ​ , ∫ 1 3/2 ​ and ∫ 3 / 2 ∞ \int_{3/2}^{\infty} ∫ 3/2 ∞ ​ are all convergent. Now show that ∫ 0 1 d x 2 x 2 − 5 x + 3 \int_0^{1}\frac{dx}{2x^2-5x+3} ∫ 0 1 ​ 2 x 2 − 5 x + 3 d x ​ is divergent ! Conclusion: ∫ 0 ∞ d x 2 x 2 − 5 x + 3 \int_0^{\infty}\frac{dx}{2x^2-5x+3} ∫ 0 ∞ ​ 2 x 2 − 5 x + 3 d x ​ is divergent.