Limit at infinity via L'Hopital's Rule

As $x \to \infty$, both the numerator $3x^2+2x \to \infty$ and the denominator $5x^2 - 7 \to \infty$, so the limit has the indeterminate form $\infty/\infty$, and L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\frac{d}{dx}(3x^2+2x) = 6x+2$ and $\frac{d}{dx}(5x^2-7) = 10x$, so the limit becomes $\lim_{x\to\infty} \dfrac{6x+2}{10x}$.

This new limit is still of the form $\infty/\infty$, so L'Hopital's Rule is applied again: differentiating gives $\frac{d}{dx}(6x+2) = 6$ and $\frac{d}{dx}(10x) = 10$, yielding $\lim_{x\to\infty} \dfrac{6}{10} = \dfrac{3}{5}$. Therefore $\lim_{x \to \infty} \dfrac{3x^2+2x}{5x^2-7} = \dfrac{3}{5}$.