Insert an Element at a Specific Index in a List and Return the Updated List

l.insert(index, obj) doesn't actually return anything. It just updates the list.

As ATO said, you can do b = a[:index] + [obj] + a[index:]. However, another way is:

a = [1, 2, 4]
b = a[:]
b.insert(2, 3)

Most performance efficient approach

You may also insert the element using the slice indexing in the list. For example:

>>> a = [1, 2, 4]
>>> insert_at = 2  # Index at which you want to insert item

>>> b = a[:]   # Created copy of list "a" as "b".
               # Skip this step if you are ok with modifying the original list

>>> b[insert_at:insert_at] = [3]  # Insert "3" within "b"
>>> b
[1, 2, 3, 4]

For inserting multiple elements together at a given index, all you need to do is to use a list of multiple elements that you want to insert. For example:

>>> a = [1, 2, 4]
>>> insert_at = 2   # Index starting from which multiple elements will be inserted

# List of elements that you want to insert together at "index_at" (above) position
>>> insert_elements = [3, 5, 6]

>>> a[insert_at:insert_at] = insert_elements
>>> a   # [3, 5, 6] are inserted together in `a` starting at index "2"
[1, 2, 3, 5, 6, 4]

To know more about slice indexing, you can refer: Understanding slice notation.

Note: In Python 3.x, difference of performance between slice indexing and list.index(...) is significantly reduced and both are almost equivalent. However, in Python 2.x, this difference is quite noticeable. I have shared performance comparisons later in this answer.

Alternative using list comprehension (but very slow in terms of performance):

As an alternative, it can be achieved using list comprehension with enumerate too. (But please don't do it this way. It is just for illustration):

>>> a = [1, 2, 4]
>>> insert_at = 2

>>> b = [y for i, x in enumerate(a) for y in ((3, x) if i == insert_at else (x, ))]
>>> b
[1, 2, 3, 4]