Limit of a linear function over a quadratic
Evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{500x + 10^4}{0.001x^2 + 1}. $$
Explanation of Method
Both numerator and denominator grow without bound as $x\to+\infty$, giving $\dfrac{\infty}{\infty}$. Since the denominator's highest power of $x$ is $x^2$ while the numerator's is only $x^1$, dividing every term — numerator and denominator alike — by $x^2$ turns every term into either a constant or a term that vanishes as $x\to+\infty$, except that the numerator's leading term vanishes too (since its power, $x$, is lower than $x^2$), leaving a nonzero constant only in the denominator.
- Divide every term of the numerator and denominator by $x^2$, the highest power present in either.
- Evaluate the limit of each resulting term as $x \to +\infty$.
- Combine the results to find $L$.
Step 1 · Divide every term by $x^2$
Dividing both the numerator and denominator by $x^2$ (valid for $x\neq0$):
$$ \frac{500x+10^4}{0.001x^2+1} = \frac{\dfrac{500x}{x^2}+\dfrac{10^4}{x^2}}{\dfrac{0.001x^2}{x^2}+\dfrac{1}{x^2}} = \frac{\dfrac{500}{x}+\dfrac{10^4}{x^2}}{0.001+\dfrac{1}{x^2}}. $$
Step 2 · Take the limit of each term
As $x \to +\infty$, every term with $x$ in the denominator vanishes:
$$ \frac{500}{x}\to0, \qquad \frac{10^4}{x^2}\to0, \qquad \frac{1}{x^2}\to0. $$
So the numerator of the simplified expression tends to $0+0=0$, while the denominator tends to $0.001+0=0.001$.
Step 3 · Combine the results
Substituting these limits:
$$ L = \frac{0}{0.001} = 0. $$
Summary Table
| Term (after dividing by $x^2$) | Limit as $x \to +\infty$ |
|---|---|
| $\dfrac{500}{x}$ | $0$ |
| $\dfrac{10^4}{x^2}$ | $0$ |
| $\dfrac1{x^2}$ | $0$ |
| Full ratio | $\dfrac{0}{0.001}=0$ |
Additional Notes
- This is the general rule for rational functions at infinity: when the denominator's degree exceeds the numerator's degree, the limit at $\pm\infty$ is always $0$, regardless of the specific coefficients involved (here, $500$, $10^4$, $0.001$, and $1$ all become irrelevant once the degree comparison is made).
- Focusing only on the "dominant terms" ($500x$ over $0.001x^2$) is a useful shortcut for quickly guessing the answer, but dividing through by the full highest power (as done here) is the rigorous version — it shows explicitly that every neglected term vanishes rather than just asserting it.
- This same rational function has a vertical behavior worth noting for context: since $0.001x^2+1 > 0$ for all real $x$, the function has no vertical asymptotes at all — its only interesting long-run behavior is the horizontal asymptote $y=0$ found here.
Result
The value of the limit is: