<s>[INST] Calculating limit $\lim\limits_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$ for an unknown function. Given that $f(x)$ is a continuous function and satisfies $f'(x)>0$ on $(-\infty,\infty)$ and $f''(x)=2 \forall x \in(0,\infty)$.We need to find the limit 
$$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)}$$ 
Now the numerator is tending to infinity so denominator must also go to infinity else limit won't exist.So I tried the L'Hospitals rule and it became$$\lim_{x\to\infty}\frac{6x+\frac{6x}{(x^2+1)^2}-4f''(x)}{f'(x)}$$The numerator is still infinity so once again applying L'Hospitals rule (assuming denominator must still be infinity) we get 
$$\lim_{x\to\infty}\frac{6+\frac{6(x^2+1)^2-6x×2(x^2+1)×2x}{(x^2+1)^4}+0}{f''(x)}$$
Now putting $f''(x)=2$ we get 
$$3+\lim_{x\to\infty}\frac{3(x^2+1)^2-12x^2(x^2+1)^2}{(x^2+1)^4}$$

Collecting the coefficients of $x^4$ from numerator and denominator we get the limit to be$3-9=-6$ but the answer is not -6. Is applying LHospital wrong?Help.Thanks.

 [/INST] Since $f''(x)=2~~for ~~x>0$, Therefore $f$ has the form $f(x)=x^2+bx +c~~~for~~~x>0$


Since the limit is at $+\infty$ it suffices to consider $x>0$
  $$\lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-4f'(x)}{f(x)} = \lim_{x\to\infty}\frac{3x^2-\frac{3}{x^2+1}-8x-4b}{x^2+bx +c} =3$$


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