Find the limit of (2x - 3)/(x^2 + 3x) as x approaches infinity.

Step-by-step solution

1. 1

   Identify the degrees before computing. The numerator has degree $1$, the denominator degree $2$. For a rational function, when the numerator degree is lower, the limit at infinity is $0$; equal degrees give the ratio of leading coefficients; a higher numerator degree gives $\pm\infty$. This problem is the first case, and the work below simply demonstrates it.

2. 2

   Divide numerator and denominator by the highest power in the denominator, $x^{2}$.

   $\frac{2x-3}{x^{2}+3x}=\frac{\frac{2x}{x^{2}}-\frac{3}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{3x}{x^{2}}}=\frac{\frac{2}{x}-\frac{3}{x^{2}}}{1+\frac{3}{x}}$

   Dividing by $x^2$ (not by $x$) is what guarantees the denominator settles on a nonzero constant.

3. 3

   Send each reciprocal power to zero. As $x\to\infty$,

   $\frac{2}{x}\to 0,\qquad\frac{3}{x^{2}}\to 0,\qquad\frac{3}{x}\to 0$

4. 4

   Assemble the limit.

   $\frac{0-0}{1+0}=\frac{0}{1}=0$

   The denominator tending to $1$ rather than $0$ is what makes this a legitimate evaluation rather than another indeterminate form.

5. 5

   Interpret the result graphically. The limit says the curve $y=\dfrac{2x-3}{x^{2}+3x}$ has the horizontal asymptote $y=0$. Numerically the values are $0.00199$ at $x=1000$ and $0.0000020$ at $x=10^{6}$, decaying like $2/x$ ✓.