Tomamos como el límite
$$\lim_{x \to \frac{201}{100}^+}\left(\frac{6 x^{3} + 9 x^{2}}{3 x + 1}\right)$$
cambiamos
$$\lim_{x \to \frac{201}{100}^+}\left(\frac{6 x^{3} + 9 x^{2}}{3 x + 1}\right)$$
=
$$\lim_{x \to \frac{201}{100}^+}\left(\frac{3 x^{2} \left(2 x + 3\right)}{3 x + 1}\right)$$
=
$$\lim_{x \to \frac{201}{100}^+}\left(\frac{x^{2} \left(6 x + 9\right)}{3 x + 1}\right) = $$
$$\frac{\left(\frac{201}{100}\right)^{2} \left(9 + \frac{6 \cdot 201}{100}\right)}{1 + \frac{3 \cdot 201}{100}} = $$
= 42542253/3515000
Entonces la respuesta definitiva es:
$$\lim_{x \to \frac{201}{100}^+}\left(\frac{6 x^{3} + 9 x^{2}}{3 x + 1}\right) = \frac{42542253}{3515000}$$