retry: Default MaxBackoff is unintuitive

Issue

I would expect the following

func Test(t *testing.T) {
	err:=retry.Do(
		context.Background(),
		func() error {
			fmt.Println(time.Now())
			return errors.New("some errors")
		},
		retry.WithInitialBackoff(10*time.Second), retry.WithMultiplier(1), retry.WithMaxAttempts(5),
	)
	require.NoError(t,err)
}

To output one log line every 10 seconds. However what this actually outputs is

2019-01-16 14:51:16.025567248 -0800 PST m=+0.011095106
2019-01-16 14:51:18.092852159 -0800 PST m=+2.078377531
2019-01-16 14:51:20.357685855 -0800 PST m=+4.343208502
2019-01-16 14:51:22.459165577 -0800 PST m=+6.444685696
2019-01-16 14:51:24.425426817 -0800 PST m=+8.410944570

The reason for this output is the default for MaxBackoff is as follows:

// WithMaxBackoff sets upper limit on backoff duration.
//
// Max backoff of 0 indicates no limit.
//
// If max backoff option is not used, then default value of 2 seconds is used.

While the documentation is good, this caused us to write bugs that affected us in production as I think it's unintuitive.

Proposal

Set the default value of max backoff to 0 instead of 2 seconds, so that the max backoff will be unlimited.

I realize this is a breaking change, but I would believe that most people using this library would expect it to work this way

Start at retry.Do and the WithMaxBackoff option described in the issue, then inspect how the default is selected and how backoff timing is tested. Review the unresolved discussion before changing this breaking behavior, and ensure coverage confirms that omitting MaxBackoff allows unlimited backoff while explicit limits remain unchanged.