In modern browsers, jQuery makes use of `document.querySelectorAll()` to boost performance when valid CSS selectors are used. It falls back to Sizzle if a browser doesn't support the selector or the `document.querySelectorAll()` method.

However, I'd like to always use Sizzle instead of the native implementation when debugging a custom selector. Namely, I'm trying to come up with an implementation of `:nth-last-child()`, [one of the CSS3 selectors that are not supported by jQuery](https://stackoverflow.com/questions/11745274/what-css3-selectors-does-jquery-really-support-e-g-nth-last-child). Since this selector is natively supported by modern browsers, it works as described in the linked question. It is precisely this behavior that's interfering with debugging my custom selector, though, so I'd like to avoid it.

A cheap hack I can use is to drop in a non-standard [jQuery selector extension](http://api.jquery.com/category/selectors/jquery-selector-extensions), thereby "invalidating" the selector so to speak. For example, assuming every `li:nth-last-child(2)` is [visible](http://api.jquery.com/visible-selector), I can simply drop that in, turning this:

```
$('li:nth-last-child(2)').css('color', 'red');

```

Into this:

```
$('li:nth-last-child(2):visible').css('color', 'red');

```

This causes it to always be evaluated by Sizzle. Except, this requires that I make an assumption of my page elements which may or may not be true. And I really don't like that. Not to mention, I dislike using non-standard selectors in general unless absolutely necessary.

Is there a way to skip the native `document.querySelectorAll()` method in browsers that support it and force jQuery to use Sizzle to evaluate a selector instead, that preferably *doesn't* employ the use of non-standard selectors? Likely, this entails calling another method instead of `$()`, but it's much better than a selector hack IMO.