Evaluate the limit of a rational function as x goes to infinity

\displaystyle\lim_{x \to \infty} \dfrac{\text{polynomial}}{\text{polynomial}}

Remember: Divide by the bottom's top power.

See it in a problem3 worked examples

\displaystyle\lim_{x \to \infty} \dfrac{4x^3 - x + 2}{1 - 2x^3}

1
$x$ $x^3$
2
$Divide every term, top and bottom, by that power$ $\dfrac{4 - \frac{1}{x^2} + \frac{2}{x^3}}{\frac{1}{x^3} - 2}$
3
$\dfrac{c}{x^n}$ $\dfrac{1}{x^2}, \dfrac{2}{x^3}, \dfrac{1}{x^3} \to 0$
4
$0$ $\dfrac{4}{-2} = -2$

-2

Common mistakes3

Saying $\dfrac{\infty}{\infty} = 1$ . It is a form, not a value.
Forgetting the sign at $-\infty$ : $\dfrac{2x^3}{7x^2} = \dfrac{2x}{7} \to -\infty$ there, not $+\infty$ .
Dividing top and bottom by different powers. Use the same power everywhere.

2

Find a limit with exponentials, powers and logs at infinity

\displaystyle\lim_{t \to \pm\infty}

Remember: Logs lose to powers, powers lose to exponentials — but check the direction.

See it in a problem4 worked examples

\displaystyle\lim_{t \to \infty} \dfrac{2e^{4t} + 3t}{5e^{3t} + t^2}

1
$List the terms on top and on bottom$ $2e^{4t}$
2
$Decide what each term does in this direction$ $t \to \infty$
3
$Keep the dominant term on top and on bottom$ $\approx 2e^{4t}$
4
$Simplify that ratio and take the limit$ $\dfrac{2e^{4t}}{5e^{3t}} = \dfrac{2}{5}e^{t} \to \infty$

\infty

Common mistakes3

Keeping $e^{x}$ as $x \to -\infty$ . It goes to $0$ there, so the polynomial terms win.
Treating $(0.6)^{x}$ like $2^{x}$ : a base below $1$ decays as $x \to \infty$ and grows without bound as $x \to -\infty$ .
Picking the dominant term of the whole fraction instead of the top and bottom separately.

3

\displaystyle\lim_{x \to \pm\infty} \dfrac{\sqrt{9x^2 + 4x}}{2x - 1}

1
$Pull the highest power out of the square root$ $\sqrt{9x^2 + 4x} = \sqrt{x^2}\,\sqrt{9 + \frac{4}{x}}$
2
$\sqrt{x^2} = |x|$ $x \to \infty$
3
$Factor the same power out of the rest; cancel$
4
$\dfrac{c}{x^n}$

Remember: Root of x squared is the absolute value.

See it in a problem3 worked examples

\displaystyle\lim_{x \to \infty} \dfrac{\sqrt{9x^2 + 4x}}{2x - 1}

1
$Pull the highest power out of the square root$ $\sqrt{9x^2 + 4x} = \sqrt{x^2}\,\sqrt{9 + \frac{4}{x}}$
2
$\sqrt{x^2} = |x|$ $x > 0$
3
$Factor the same power out of the rest; cancel$ $\dfrac{x\sqrt{9 + \frac{4}{x}}}{x\left(2 - \frac{1}{x}\right)} = \dfrac{\sqrt{9 + \frac{4}{x}}}{2 - \frac{1}{x}}$
4
$\dfrac{c}{x^n}$ $\dfrac{\sqrt{9}}{2} = \dfrac{3}{2}$

\dfrac{3}{2}

Common mistakes2

Writing $\sqrt{x^2} = x$ as $x \to -\infty$ . The root is positive, $x$ is negative: the answer flips sign.
Pulling $x^2$ out of the root but only $x$ out of one term inside it. Every term inside gets divided by $x^2$ .

4

\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + 8x} - x\right)

Remember: Conjugate, cancel the squares, divide by x.

See it in a problem3 worked examples

\displaystyle\lim_{x \to \infty} \left(\sqrt{x^2 + 8x} - x\right)

1
$\infty - \infty$ $\sqrt{x^2 + 8x} \to \infty$
2
$Multiply and divide by the conjugate$ $\left(\sqrt{x^2 + 8x} - x\right) \cdot \dfrac{\sqrt{x^2 + 8x} + x}{\sqrt{x^2 + 8x} + x}$
3
$A - B^2$ $(x^2 + 8x) - x^2 = 8x$
4
$x$ $x$
5
$\dfrac{c}{x^n}$ $\dfrac{8}{1 + 1} = 4$

4

Common mistakes2

Saying $\infty - \infty = 0$ . Two things that both grow without bound can differ by any amount.
Using $\sqrt{x^2} = x$ when $x \to -\infty$ ; there it is $-x$ .

5

f(x)

Remember: Big x: biggest power. Small x: smallest power.

See it in a problem4 worked examples

f(x) = \dfrac{x^4 - 2x^7}{3 + x^3}

f(x) \approx -2x^4

Common mistakes3

Using "highest power" near $0$ . There $x^7$ is far smaller than $x^4$ , so the lowest power wins.
Dropping the coefficient: $\dfrac{x^4}{3}$ , not $x^4$ . Keep the constants of the kept terms.
Forgetting $e^{t} \to 0$ as $t \to -\infty$ : there the power terms take over.