After some consideration, in my opinion, "lower boundary" will make more sense rather than lower limit.
For example, this is the data,
Class Frequency
1 1
2 1
3 1
4 1
Based on the data, using we can know that the median is 2.5, without calculation. If using the formula as mentioned above, $\frac{n}{2}$ will get 2, there for the class contains the median is class 2, then using $L_m$ is a lower boundary,
$median = 1.5 + \left[ \frac{2 -1}{1}\right] \times 1 = 2.5$
This doesn't make sense for using lower limit. If changing the class to
Class Frequency
1-2 1
3-4 1
5-6 1
7-8 1
Using the method above, we will get,
$median = 2.5 + \left[ \frac{2 -1}{1}\right] \times 2 = 4.5$
However, if using class limit, then we will get 5.
$$
\mathrm{Median} = L_m + \frac{n/2 -cf_b}{f_m}*w
$$
here
$L_m$ = median group where the median lies in
$n$ = total number of data
$cf_b$ = culm. freq before the median group
$f_m$ = frequency of median group
$w$ = width of groups
So for your problem:
$L_m$ = (6)-10,$n$ = 496, $cf_b$ = 100, $f_m$ = 340 and $w$ = 5
This leads to:
$$
\mathrm{Median} = 6 + \frac{496/2-100}{340}5 = 6 + \frac{148*5}{340} = 6 + 2.176 \approx 8.18
$$
So the error was choosing the right numbers for the formula, especially where the median lies.
Best Answer
This formula is the result of a linear interpolation, which identifies the median under the assumption that data are uniformly distributed within the median class.
To derive the formula, we can note that since $N/2$ is the number of observations below the median, then $N/2 - F_{m-1}$ is the number of observations that are within the median class and that are below the median ($F_{m-1}$ is the cumulative frequency of the interval below the median class, i.e. of all classes below the median class).
As a result, the fraction $\displaystyle\frac {N/2 - F_{m-1}}{f_m}$ (where $f_m$ is the frequency of the median class) represents the proportion of data values in the median class that are below the median.
Now if we assume that data are uniformly distributed (i.e., equally spaced) within the median class, multiplying the last fraction by $c$ (total width of the median class) we obtain the fraction of median class width corresponding to the position of the median. Adding the result to $L_m$ (lower limit of the median class), we get the final formula $\displaystyle L_m + \left [ \frac { \frac{N}{2} - F_{m-1} }{f_m} \right ] \cdot c$, which identifies the median.