# Calculus: Limits

## Limits at Infinity

Limits are not limited to being taken at real numbers. Taking the limit of a function as it tends towards positive infinity, $\infty$, or negative infinity, $-\infty$, is also an interesting thing to do. Does this function tend towards the infinite? Or does it settle on a single value?

However, limits at infinity need not converge to a single value. Functions that oscillate, such as sine and cosine, neither converge to a single value nor grow indefinitely towards either $\infty$ or $-\infty$. Since they just keep repeating, asking where they might "end" is a nonsensical thing to ask.

As for solving limits at infinity, two additional Limit Laws will be of use:

1. If $\lim\limits_{x \rightarrow \infty} f(x) = \infty$, then $\lim\limits_{x \rightarrow \infty} \dfrac{1}{f(x)} = 0$.
2. The limits of periodic functions, such as sine, cosine, and tangent, as $x$ goes towards positive or negative infinity do not exist.

## Problems

1. Let $f(x) = \dfrac{1}{x^2+1}$. Find $\lim\limits_{x \rightarrow \infty} \, f(x)$ using the following techniques:

1. Graphing
2. Substitution
1. Graphing:

By looking at the graph, we can reasonably conclude that the function converges to $0$. In math:

$\lim\limits_{x \rightarrow \infty} \dfrac{1}{x^2+1} = 0$

2. Substitution:
$\lim\limits_{x \rightarrow \infty} f(x) = \lim\limits_{x \rightarrow \infty} \dfrac{1}{x^2+1} \\ = \dfrac{1}{\infty^2 + 1} \\ = \dfrac{1}{\infty} \\ = 0$
2. Find the limit: $\lim\limits_{x \rightarrow \infty} \dfrac{1}{x} + e^{1/x} + 2e^{-x}$

$\lim\limits_{x \rightarrow \infty} \dfrac{1}{x} + e^{1/x} + 2e^{-x} = \dfrac{1}{\infty} + e^{1/\infty} + 2e^{-\infty} \\ = 0 + e^{0} + 2\cdot 0 \\ = 1$
3. Find the limit: $\lim\limits_{x \rightarrow -\infty} \pi^{1/x} -\dfrac{2}{x} - 3e^{-x}$
$\lim\limits_{x \rightarrow -\infty} \pi^{1/x} -\dfrac{2}{x} - 3e^{-x} = \pi^{1/-\infty} -\dfrac{2}{-\infty} - 3e^{-(-\infty)} \\ = \dfrac{1}{\pi^{1/\infty}} + \dfrac{2}{\infty}- 3e^{\infty} \\ = \dfrac{1}{\pi^{0}} + 0 - \infty \\ = \dfrac{1}{1} + 0 - \infty \\ = -\infty$

The limit does not exist.
4. Find the limit: $\lim\limits_{x \rightarrow \infty} \dfrac{e^x + 16}{e^x}$

First try direct subtitution: $\lim\limits_{x \rightarrow \infty} \dfrac{e^x + 16}{e^x} = \dfrac{e^{\infty} + 16}{e^{\infty}} \\ = \dfrac{\infty}{\infty}$

This is an indeterminate form. Separating the terms should solve this issue:

$\lim\limits_{x \rightarrow \infty} \dfrac{e^x + 16}{e^x} = \dfrac{e^x}{e^x} + \dfrac{16}{e^x} \\ = 1 + \dfrac{16}{e^x} \\ = 1 + \dfrac{16}{e^{\infty}} \\ = 1 + \dfrac{16}{\infty} \\ = 1 + 0 \\ = 1$
5. Find the limit: $\lim\limits_{x \rightarrow \infty} x^3 + e^x + 2$

$\lim\limits_{x \rightarrow \infty} x^3 + e^x + 2 = \infty^3 + e^{\infty} + 2 \\ = \infty + \infty + 2 \\ = \infty$

The limit does not exist.
6. Find the limit: $\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 2$

First try direct substitution:

$\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 2 = (-\infty)^4 - (-\infty)^2 - 2 \\ = \infty - \infty - 2 \\ = \infty - \infty$

This is an indeterminate form. We can get around this by factoring the expression first:

$\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 2 = \lim\limits_{x \rightarrow -\infty} (x^2 - 2)(x^2 + 1) \\ = (\infty^2 - 2)(\infty^2 + 1) \\ = \infty \cdot \infty \\ = \infty$

The limit does not exist.
7. Find the limit: $\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 3$

First try direct substitution:

$\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 3 = (-\infty)^4 - (-\infty)^2 - 3 \\ = \infty - \infty - 3 \\ = \infty - \infty$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow -\infty} x^4 - x^2 - 3 = \lim\limits_{x \rightarrow -\infty} x^4\left(1 - \dfrac{1}{x^2} - \dfrac{3}{x^4} \right) \\ = \infty^4\left(1 - \dfrac{1}{(-\infty)^2} - \dfrac{3}{(-\infty)^4} \right) \\ = \infty(1 - 0 - 0 ) \\ = \infty$

The limit does not exist.
8. Find the limit: $\lim\limits_{x \rightarrow \infty} \sin(x^{-x})$

$\lim\limits_{x \rightarrow \infty} \sin(x^{-x}) = \sin\left(\dfrac{1}{x^{x}}\right) \\ = \sin\left(\dfrac{1}{\infty^{\infty}}\right) \\ = \sin(0) \\ = 0$
9. Find the limit: $\lim\limits_{x \rightarrow -\infty} \dfrac{2x^3 - 4}{x^3 + 4x}$

First try direct substitution:

$\lim\limits_{x \rightarrow -\infty} \dfrac{2x^3 - 4}{x^3 + 4x} = \dfrac{2\infty^3 - 4}{\infty^3 + 4\infty} \\ = \dfrac{\infty}{\infty + \infty} \\ = \dfrac{\infty}{\infty}$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow -\infty} \dfrac{2x^3 - 4}{x^3 + 4x} = \lim\limits_{x \rightarrow -\infty} \dfrac{x^3\left(2 - \frac{4}{x^3}\right)}{x^3\left(1 + \frac{4}{x^2}\right)} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{2 - \frac{4}{x^3}}{1 + \frac{4}{x^2}} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{2 - \frac{4}{(-\infty)^3}}{1 + \frac{4}{\infty^2}} \\ = \dfrac{2 + 0}{1 + 0} \\ = 2$
10. Find the limit: $\lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 2x + 2}{x^2 + 4x + 40}$

First try direct substitution:

$\lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 2x + 2}{x^2 + 4x + 40} = \dfrac{\infty^2 + 2\infty + 2}{\infty^2 + 4\infty + 40} \\ = \dfrac{\infty + \infty }{\infty + \infty} \\ = \dfrac{\infty}{\infty}$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 2x + 2}{x^2 + 4x + 40} = \lim\limits_{x \rightarrow -\infty} \dfrac{x^2\left(1 + \frac{2}{x} + \frac{2}{x^2}\right)}{x^2\left(1 + \frac{4}{x} + \frac{40}{x^2}\right)} \\ = \dfrac{1 + \frac{2}{\infty} + \frac{2}{\infty^2}}{1 + \frac{4}{\infty} + \frac{40}{\infty^2}} \\ = \dfrac{1 + 0 + 0}{1 + 0 + 0} \\ = 1$
11. Find the limit: $\lim\limits_{x \rightarrow -\infty} \dfrac{7x}{3x^2 - x + 3}$

First try direct substitution:

$\lim\limits_{x \rightarrow -\infty} \dfrac{7x}{3x^2 - x + 3} = \lim\limits_{x \rightarrow -\infty} \dfrac{7\infty}{3(-\infty)^2 + \infty + 3} \\ = \dfrac{\infty}{\infty + \infty} \\ = \dfrac{\infty}{\infty}$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow -\infty} \dfrac{7x}{3x^2 - x + 3} = \lim\limits_{x \rightarrow -\infty} \dfrac{x^2\left(\frac{7}{x}\right)}{x^2\left(3 - \frac{1}{x} + \frac{3}{x^2}\right)} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{\frac{7}{x}}{3 - \frac{1}{x} + \frac{3}{x^2}} \\ = \dfrac{\frac{7}{-\infty}}{3 + \frac{1}{\infty} + \frac{3}{(-\infty)^2}} \\ = \dfrac{0}{3 + 0 + 0} \\ = 0$
12. Find the limit: $\lim\limits_{x \rightarrow \infty} \dfrac{x^5}{x^4 + x^2 + x}$

First try direct substitution:

$\lim\limits_{x \rightarrow \infty} \dfrac{x^5}{x^4 + x^2 + x} = \lim\limits_{x \rightarrow \infty} \dfrac{\infty^5}{\infty^4 + \infty^2 + \infty} \\ = \dfrac{\infty}{\infty + \infty + \infty} \\ = \dfrac{\infty}{\infty}$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow \infty} \dfrac{x^5}{x^4 + x^2 + x} = \lim\limits_{x \rightarrow \infty} \dfrac{x^5}{x^5\left(\frac{1}{x} + \frac{1}{x^3} + \frac{1}{x^4}\right)} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{1}{\frac{1}{x} + \frac{1}{x^3} + \frac{1}{x^4}} \\ = \dfrac{1}{\frac{1}{\infty} + \frac{1}{\infty^3} + \frac{1}{\infty^4}} \\ = \dfrac{1}{0 + 0 + 0} \\ = \infty$

The limit does not exist.
13. Find the limit: $\lim\limits_{x \rightarrow \infty} \dfrac{x^4 - 19x}{9x^4 - 18}$

First try direct substitution:

$\lim\limits_{x \rightarrow \infty} \dfrac{x^4 - 19x}{9x^4 - 18} = \lim\limits_{x \rightarrow \infty} \dfrac{\infty^4 - 19\infty}{9\infty^4 - 18} \\ = \dfrac{\infty - \infty}{\infty } \\$

This is an indeterminate form. We can get around this by factoring out the leading polynomial term:

$\lim\limits_{x \rightarrow \infty} \dfrac{x^4 - 19x}{9x^4 - 18} = \lim\limits_{x \rightarrow \infty} \dfrac{x^4\left(1 - \frac{19}{x^3}\right)}{x^4\left(9 - \frac{18}{x^4}\right)} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{1 - \frac{19}{x^3}}{9 - \frac{18}{x^4}} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{1 - \frac{19}{\infty^3}}{9 - \frac{18}{\infty^4}} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{1 - 0}{9 - 0} \\ = \dfrac{1}{9} \\$
14. Find the limit: $\lim\limits_{x \rightarrow \infty} x - \sqrt{ x^2 - 4 }$

First try direction substitution:

$\lim\limits_{x \rightarrow \infty} x - \sqrt{ x^2 - 4 } = \infty - \sqrt{\infty^2 - 4} \\ = \infty - \sqrt{\infty} \\ = \infty - \infty$

This is an indeterminate form. The solution to this is to multiply by $1$ in a special way, namely by multiplying by the conjugate over the conjugate:

$\lim\limits_{x \rightarrow \infty} x - \sqrt{ x^2 - 4 } = \lim\limits_{x \rightarrow \infty}(x - \sqrt{ x^2 - 4 }) \dfrac{x + \sqrt{ x^2 - 4 }}{x + \sqrt{ x^2 - 4 }} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{x^2 - x^2 - 4 }{x + \sqrt{ x^2 - 4 }} \\ = \lim\limits_{x \rightarrow \infty} \dfrac{-4}{x + \sqrt{ x^2 - 4 }} \\ = \dfrac{-4}{\infty + \sqrt{ \infty^2 - 4 }} \\ = \dfrac{-4}{\infty + \sqrt{\infty}} \\ = \dfrac{-4}{\infty} \\ = 0$
15. Find the limit: $\lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 9}{\sqrt{x^4 - 9}}$

Notice that plugging in $-\infty$ directly will lead to the indeterminate form of $\dfrac{\infty}{\infty}$. The solution is to rationalize the denominator:

$\lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 9}{\sqrt{x^4 - 9}} = \lim\limits_{x \rightarrow -\infty} \dfrac{x^2 + 9}{\sqrt{x^4 - 9}}\left(\dfrac{\sqrt{x^4 - 9}}{\sqrt{x^4 - 9}}\right) \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{(x^2+9)\sqrt{x^4 - 9}}{x^4 - 9} \\$

Motice again that plugging in $-\infty$ will lead to the same indeterminate form of $\dfrac{\infty}{\infty}$. The solution is to keep factoring out leading polynomial terms until we reach a determinate form:

$= \lim\limits_{x \rightarrow -\infty} \dfrac{(x^2+9)\sqrt{x^4\left(1 - \frac{9}{x^4}\right)}}{x^4 - 9} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{(x^2+9)x^2\sqrt{1 - \frac{9}{x^4}}}{x^4 - 9} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{(x^4+9x^2)\sqrt{1 - \frac{9}{x^4}}}{x^4 - 9} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{x^4\left(1 + \frac{9}{x^2}\right)\sqrt{1 - \frac{9}{x^4}}}{x^4\left(1 - \frac{9}{x^4}\right)} \\ = \lim\limits_{x \rightarrow -\infty} \dfrac{\left(1 + \frac{9}{x^2}\right)\sqrt{1 - \frac{9}{x^4}}}{1 - \frac{9}{x^4}} \\ = \dfrac{\left(1 + \frac{9}{(-\infty)^2}\right)\sqrt{1 - \frac{9}{(-\infty)^4}}}{1 - \frac{9}{(-\infty)^4}} \\ = \dfrac{\left(1 + 0\right)\sqrt{1 - 0}}{1 - 0} \\ = 1$