doc: clarify that fs.exists() may return false for existing symlink

Given that this API is problematic in any case, we should be precise
about its (perhaps surprising) behavior.

PR-URL: #53566
Reviewed-By: Jithil P Ponnan <[email protected]>
Reviewed-By: Moshe Atlow <[email protected]>
Reviewed-By: Yagiz Nizipli <[email protected]>
Reviewed-By: Marco Ippolito <[email protected]>
Reviewed-By: Luigi Pinca <[email protected]>
Reviewed-By: Ulises Gascón <[email protected]>

Author: tniessen

doc/api/fs.md

@@ -2656,7 +2656,7 @@ changes:
 * `callback` {Function}
   * `exists` {boolean}
 
-Test whether or not the given path exists by checking with the file system.
+Test whether or not the element at the given `path` exists by checking with the file system.
 Then call the `callback` argument with either true or false:
 
 ```mjs
@@ -2673,6 +2673,9 @@ parameter, optionally followed by other parameters. The `fs.exists()` callback
 has only one boolean parameter. This is one reason `fs.access()` is recommended
 instead of `fs.exists()`.
 
+If `path` is a symbolic link, it is followed. Thus, if `path` exists but points
+to a non-existent element, the callback will receive the value `false`.
+
 Using `fs.exists()` to check for the existence of a file before calling
 `fs.open()`, `fs.readFile()`, or `fs.writeFile()` is not recommended. Doing
 so introduces a race condition, since other processes may change the file's