a company training program has determined that, on the average, a new employee produces ( p(s) ) items per…

a company training program has determined that, on the average, a new employee produces ( p(s) ) items per day after ( s ) days of on - the - job training, where ( p(s)=\frac{68 s}{s + 5} ). find and interpret ( lim _{s \rightarrow infty} p(s) ). find the limit. select the correct choice below and, if necessary, fill in the answer box within your choice. a. ( lim _{s \rightarrow infty} p(s)= ) b. the limit does not exist and is not ( infty ) or ( -infty ).

a company training program has determined that, on the average, a new employee produces ( p(s) ) items per day after ( s ) days of on - the - job training, where ( p(s)=\frac{68 s}{s + 5} ). find and interpret ( lim _{s \rightarrow infty} p(s) ). find the limit. select the correct choice below and, if necessary, fill in the answer box within your choice. a. ( lim _{s \rightarrow infty} p(s)= ) b. the limit does not exist and is not ( infty ) or ( -infty ).

Answer

Explanation:

Step1: Divide numerator and denominator by (s)

Divide each term in (P(s)=\frac{68s}{s + 5}) by (s). We get (\frac{\frac{68s}{s}}{\frac{s}{s}+\frac{5}{s}}=\frac{68}{1+\frac{5}{s}})

Step2: Apply the limit

We know that (\lim_{s\rightarrow\infty}\frac{1}{s}=0). So, (\lim_{s\rightarrow\infty}\frac{68}{1+\frac{5}{s}}=\frac{\lim_{s\rightarrow\infty}68}{\lim_{s\rightarrow\infty}(1 + \frac{5}{s})}) Since (\lim_{s\rightarrow\infty}68 = 68) and (\lim_{s\rightarrow\infty}(1+\frac{5}{s})=\lim_{s\rightarrow\infty}1+\lim_{s\rightarrow\infty}\frac{5}{s}=1 + 0=1)

Answer:

(\lim_{s\rightarrow\infty}P(s)=68). This means that as the number of days of on - the - job training (s) becomes extremely large (approaching infinity), the average number of items produced per day by a new employee approaches (68). So, there is an upper limit of approximately (68) items per day that a new employee can be expected to produce after a very long period of on - the - job training.