Limit of (4x − 3)/(2x + 1) at infinity
Evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{4x-3}{2x+1}. $$
Explanation of Method
Both the numerator and denominator grow without bound as $x\to+\infty$, giving the indeterminate form $\dfrac{\infty}{\infty}$. The standard technique for a ratio of two linear (or, more generally, same-degree polynomial) expressions is to divide numerator and denominator by the highest power of $x$ appearing in the expression — here, just $x$ itself. This turns the lower-order terms into fractions like $\dfrac1x$, which vanish as $x\to+\infty$, leaving only the ratio of the leading coefficients.
- Factor $x$ out of both the numerator and the denominator.
- Cancel the common factor of $x$.
- Take the limit, using $\frac1x\to0$ as $x\to+\infty$.
Step 1 · Factor $x$ out of numerator and denominator
Both the numerator and denominator have degree $1$ in $x$, so factor $x$ out of each:
$$ \frac{4x-3}{2x+1} = \frac{x\left(4-\frac3x\right)}{x\left(2+\frac1x\right)}. $$
Step 2 · Cancel the common factor of $x$
For $x \neq 0$, the factor of $x$ cancels between numerator and denominator:
$$ \frac{4x-3}{2x+1} = \frac{4-\frac3x}{2+\frac1x}. $$
Step 3 · Take the limit as $x \to +\infty$
As $x \to +\infty$, both $\dfrac3x \to 0$ and $\dfrac1x \to 0$, so:
$$ L = \lim_{x \to +\infty} \frac{4-\frac3x}{2+\frac1x} = \frac{4-0}{2+0} = \frac{4}{2} = 2. $$
Summary Table
| Quantity | After dividing by $x$ | Limit as $x \to +\infty$ |
|---|---|---|
| Numerator | $4-\dfrac3x$ | $4$ |
| Denominator | $2+\dfrac1x$ | $2$ |
| Ratio | — | $2$ |
Additional Notes
- Since the limit is $\frac42=2$, the horizontal line $y=2$ is a horizontal asymptote of the graph of $\dfrac{4x-3}{2x+1}$ as $x\to+\infty$ (and, by the same computation with $x\to-\infty$, also as $x\to-\infty$).
- This "divide by the highest power" technique generalizes immediately: for any rational function where the numerator and denominator have the same degree, the limit at $\pm\infty$ is simply the ratio of their leading coefficients — here $\frac{4}{2}=2$, matching the coefficients of $x$ in the numerator and denominator.
- If the numerator's degree were higher than the denominator's, the same technique would show the limit is $\pm\infty$; if lower, the limit would be $0$ — the relative degrees of numerator and denominator fully determine the outcome.
Result
The value of the limit is: