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) ).\nfind the limit. select the correct choice below and, if necessary, fill in the answer box within your choice.\na. ( lim _{s \rightarrow infty} p(s)=68 )\nb. the limit does not exist and is not ( infty ) or ( -infty ).\ninterpret the limit. select the correct choice below and fill in the answer box within your choice.\na. the number of days of training for a new employee gets closer and closer to ( square ) as the number of items that employee produces increases.\nb. the number of days of training for a new employee gets closer and closer to ( square ) as the number of items that employee produces decreases.\nc. the number of items a new employee produces gets closer and closer to ( square ) as the number of days of training decreases.\nd. the number of items a new employee produces gets closer and closer to ( square ) as the number of days of training increases.

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) ).\nfind the limit. select the correct choice below and, if necessary, fill in the answer box within your choice.\na. ( lim _{s \rightarrow infty} p(s)=68 )\nb. the limit does not exist and is not ( infty ) or ( -infty ).\ninterpret the limit. select the correct choice below and fill in the answer box within your choice.\na. the number of days of training for a new employee gets closer and closer to ( square ) as the number of items that employee produces increases.\nb. the number of days of training for a new employee gets closer and closer to ( square ) as the number of items that employee produces decreases.\nc. the number of items a new employee produces gets closer and closer to ( square ) as the number of days of training decreases.\nd. the number of items a new employee produces gets closer and closer to ( square ) as the number of days of training increases.

Answer

Explanation:

Step1: Divide numerator and denominator by (s)

Given (P(s)=\frac{68s}{s + 5}). Divide numerator and denominator by (s): [ \begin{align*} \lim_{s\rightarrow\infty}P(s)&=\lim_{s\rightarrow\infty}\frac{\frac{68s}{s}}{\frac{s}{s}+\frac{5}{s}}\ &=\lim_{s\rightarrow\infty}\frac{68}{1+\frac{5}{s}} \end{align*} ]

Step2: Evaluate the limit

We know that (\lim_{s\rightarrow\infty}\frac{1}{s}=0). So, (\lim_{s\rightarrow\infty}\frac{5}{s}=0). Then: [ \lim_{s\rightarrow\infty}\frac{68}{1+\frac{5}{s}}=\frac{68}{1 + 0}=68 ]

For the interpretation, the limit (\lim_{s\rightarrow\infty}P(s)) means as (s) (the number of days of training) increases, (P(s)) (the number of items produced) approaches the limit value.

Answer:

A. (\lim_{s\rightarrow\infty}P(s)=68)

D. The number of items a new employee produces gets closer and closer to (68) as the number of days of training increases.