Approximation ratio of greedy algorithm for makespan
The lecture notes might be misquoting the example. Here is a correct version, copied from lecture notes of Ola Svensson:

The 4/3 bound is tight, an infinite family of instances showing this is given below.
Instance: we are given $m$ machines, and $2m+1$ jobs. There are three jobs with processing time $m$, and $2$ jobs with processing times $m+ 1,m+ 2,\ldots,2m−1$ each. In case of LPT, all but one of the machines get two jobs with a total processing time of $3m−1$, and a single machine gets three jobs with a total of $4m−1$ processing time.  Thus, the makespan is $4m−1$.  OPT schedules the three $m$ jobs on a single machine, and the remaining jobs on the remaining $m−1$ machines, such that each of those machines get jobs with a total processing time of $3m$, thus the makespan of OPT is $3m$.  As $m$ grows towards infinity the approximation ratio approaches $4/3$.

In the quote, LPT is the greedy algorithm. The approximation ratio for given $m$ is $\frac{4m-1}{3m} = \frac43 - \frac1m$, which tends to $\frac 43$ as $m\to\infty$.