> For the complete documentation index, see [llms.txt](https://solomons-mathbook.gitbook.io/calculus-basics/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://solomons-mathbook.gitbook.io/calculus-basics/comment-389065252/comment-389097947.md).

# Limits at infinity

No matter why kinds of Limits you're looking for, to understand it better, the best way is to read the `Step-by-Step Solution` from `Symbolab`: \[Limit Calculator from Symbolab.]\(<https://www.symbolab.com/solver/limit-calculator/\lim_{x\to\infty}\left(\frac{6x^{2}-x}{\sqrt{9x^{4}%2B7x^{3}}}\right>))

## `Rational functions`

> The KEY point is to look at the powers & coefficients of Numerator & Dominator. Just the same with `Finding the Asymptote`.

[Refer to previous note on the `How to find Asymptote`](https://github.com/solomonxie/solomonxie.github.io/issues/44#issuecomment-374894945).

![image](https://user-images.githubusercontent.com/14041622/40049084-5303489a-5866-11e8-9ba5-d2ffe6045955.png)

### Example

![image](https://user-images.githubusercontent.com/14041622/40048425-c4959c30-5864-11e8-947d-cd5c59375f72.png) Solve:

## `Quotients with square roots`

> The KEY point is to calculate both `numerator & dominator`, then calculate the limit of EACH term with in the square root.

### Example

![image](https://user-images.githubusercontent.com/14041622/40047707-c2a05052-5862-11e8-805a-5d8c6ca98d47.png) Solve: \[Refer to Symbolab step-by-step solution.]\(<https://www.symbolab.com/solver/limit-calculator/\lim_{x\to\infty}\left(\frac{6x^{2}-x}{\sqrt{9x^{4}%2B7x^{3}}}\right>))

* Divide by **highest** dominator power to get:

  ![image](https://user-images.githubusercontent.com/14041622/40047885-48fdaf46-5863-11e8-85b8-118c09272fbf.png)
* Calculate separately the limit of `Numerator` & `Dominator`:

  ![image](https://user-images.githubusercontent.com/14041622/40047966-896c2990-5863-11e8-924e-8898d747fafd.png)
* Calculate the `Square root`: Need to find limits for **EACH** term **inside** the square root.

  ![image](https://user-images.githubusercontent.com/14041622/40048191-2dc755d2-5864-11e8-8733-e91ca609069f.png)

  ![image](https://user-images.githubusercontent.com/14041622/40048201-330160c4-5864-11e8-9b7b-b98fbae043f5.png)

  ![image](https://user-images.githubusercontent.com/14041622/40048216-3c8970b4-5864-11e8-8496-68e2e638df1d.png)
* Then get the result easily.

## `Quotients with trig`

> The KEY point is to apply the `Squeeze theorem`, and it is a MUST.

### Example

![image](https://user-images.githubusercontent.com/14041622/40047015-e86fb63a-5860-11e8-9f72-b97d62f785c2.png) Solve:

* Know that `-1 ≦ cos(x) ≦ 1`, so we can tweak it to apply the `squeeze theorem` to get its limit.
* Make the inequality to: `3/-1 ≦ 3/cos(x)/-1 ≦ 3/1`
* Get that right side `3/-1 = -1` and left side `3/1 =1` is not equal.
* So the limit doesn't exist.

**Easier solution steps**:

* Know the inequality `-1 ≦ cos(x) ≦ 1`
* Replace `cos(x)` to `±1` in the equation, `3/±1`.
* Calculate limits of two sides.
* If the results are exactly the same, then the limit is the result; Otherwise the limit doesn't exist.

### Example

![image](https://user-images.githubusercontent.com/14041622/40047463-2181c778-5862-11e8-8658-d439cdcecf65.png) Solve:

* Know that `-1 ≦ sin(x) ≦ 1`
* Replace `sin(x)` as `±1`
* Left side becomes `(5x+1)/(x-5)`, right side becomes `(5x-1)/(x-5)`
* Both sides' limits are `5`, so the limit exists, and is `5`.