We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 2^+}\left(3 x^{2} - 7 x + 2\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 2^+}\left(2 x^{2} - 5 x + 2\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} + \left(2 - 7 x\right)}{2 x^{2} + \left(2 - 5 x\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 2^+}\left(\frac{3 x^{2} - 7 x + 2}{2 x^{2} - 5 x + 2}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{\frac{d}{d x} \left(3 x^{2} - 7 x + 2\right)}{\frac{d}{d x} \left(2 x^{2} - 5 x + 2\right)}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{6 x - 7}{4 x - 5}\right)$$
=
$$\lim_{x \to 2^+}\left(\frac{6 x - 7}{4 x - 5}\right)$$
=
$$\frac{5}{3}$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)