Ajax script for jquery 3.5 to make form filled automatically

Question

i have a problem with my ajax script. I want to make the form filled automatically by select the user_id. But it didn't work. I use jquery ver. 3.5.1 and this is my code:

<form class="form-horizontal" action="" method="post">
  <div class="form-group row">
    <div class="col-sm-8">
      <div class="form-group row">
        <label for="user_id" class="col-sm-4 col-form-label">User ID</label>
          <div class="col-sm-8">
          <select class="form-control" name="user_id" id="user_id" onchange="chcek_database()">
            <option value="" selected="">User ID</option>
            <?php 
              include "connection.php";
              $query=mysqli_query($con,"select * from customer");
                while ($data=mysqli_fetch_array($query)){
                  echo "<option value='$data[user_id]'>$data[user_id]</option>";
              }?>
           </select>
         </div>
       </div>
       <div class="form-group row">
         <label for="name" class="col-sm-4 col-form-label">Name</label>
           <div class="col-sm-8">
             <input type="text" class="form-control" name="name" id="name">
           </div>
       </div>
    </form>

and it's the ajax script

<script type="text/javascript">
function check_database(){
    var user_id = $("#user_id").val();
    $.ajax({
        url: 'ajax.php',
        data: "user_id="+user_id ,
    }).success(function (data) {
        var json = data,
        obj = JSON.parse(json);
        $('#name').val(obj.name);
    });
}

 <?php
    include 'connection.php';
    $nim = $_GET['user_id'];
    $query = mysqli_query($con, "SELECT * FROM customer WHERE user_id= '$user_id'");
    $customer = mysqli_fetch_array($query);
    $data = array(
                'name'          =>  $customer['name'],);
                
     echo json_encode($data);
    ?>

and when i try to select the user_id, the name won't shown. i try to check the console with inspect the google chrome and it's the error message said:

(index):406 Uncaught TypeError: $.ajax(...).success is not a function
    at cek_database ((index):406)
    at HTMLSelectElement.onchange ((index):61)
cek_database @ (index):406
onchange @ (index):61

How to fix it? hope you guys would to give me the answer. Thank you.