Q.Evaluate $\lim_{x \to \infty} \dfrac{3x^2+5x-1}{5x^2-2x+7}$.

As $x \to \infty$, both numerator and denominator grow without bound, giving the $\dfrac{\infty}{\infty}$ indeterminate form, resolved by dividing every term by the highest power of $x$ present.

Dividing every term by $x^2$ makes every term except the leading coefficients tend to $0$, leaving $\dfrac{3}{5}$, the ratio of the two leading coefficients (since numerator and denominator have equal degree).

As $x \to \infty$, both $3x^2+5x-1 \to \infty$ and $5x^2-2x+7 \to \infty$, so direct substitution gives the indeterminate form $\dfrac{\infty}{\infty}$. The standard technique is to divide every term of numerator and denominator by the highest power of $x$ present, which here is $x^2$: $\dfrac{3x^2+5x-1}{5x^2-2x+7} = \dfrac{3+\dfrac5x-\dfrac1{x^2}}{5-\dfrac2x+\dfrac7{x^2}}$

As $x \to \infty$, every term of the form $\dfrac{k}{x}$ or $\dfrac{k}{x^2}$ tends to $0$, leaving: $\lim_{x \to \infty} \dfrac{3+\dfrac5x-\dfrac1{x^2}}{5-\dfrac2x+\dfrac7{x^2}} = \dfrac{3+0-0}{5-0+0} = \dfrac35$

This matches the general rule: since the numerator and denominator are both degree 2 polynomials (equal degree), the limit as $x \to \infty$ is simply the ratio of their leading coefficients, $\dfrac35$.

Numerical cross-check: at $x=1{,}000$, numerator $=3{,}005{,}999$, denominator $=4{,}998{,}007$, giving a ratio $\approx 0.60128$, close to $\dfrac35=0.6$ and closing in further as $x$ grows larger, confirming the algebraic result.

Shortcut: for equal-degree polynomials, the limit at $\infty$ is the ratio of leading coefficients directly — $3/5$ — without the intermediate division step.

A common error is substituting $x=\infty$ directly into each term and treating $\infty - 1 = \infty$, $\infty+5\infty=\infty$, etc., concluding the limit is '$\infty/\infty$, which is undefined' — infinity is not a number that ordinary arithmetic can be performed on; the division-by-highest-power technique is what correctly resolves the form.