An efficiency consultant determines that when new workers are hired to wait tables at an upscale restaurant,the average number of tables they can wait on in a 6 hour shift is given by $N ( x ) = \frac { \sqrt { 5.3 x ^ { 2 } + 50 x + 14 } } { 0.1 x + 0.8 }$ where x is the number of shifts they've worked since being hired.What happens to an average waiter's productivity in the long run (as $x \rightarrow \infty$ )?
A)It approaches 53 tables per shift.
B)It increases without bound.
C)It approaches 23 tables per shift.
D)It approaches 14 tables per shift.

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The Answer of An efficiency consultant determines that when new workers are hired to wait tables at an upscale restaurant,the average number of tables they can wait on in a 6 hour shift is given by $N ( x ) = \frac { \sqrt { 5.3 x ^ { 2 } + 50 x + 14 } } { 0.1 x + 0.8 }$ where x is the number of shifts they've worked since being hired.What happens to an average waiter's productivity in the long run (as $x \rightarrow \infty$ )?
A)It approaches 53 tables per shift.
B)It increases without bound.
C)It approaches 23 tables per shift.
D)It approaches 14 tables per shift.

An Efficiency Consultant Determines That When New Workers Are Hired N(x)=5.3x2+50x+140.1x+0.8N ( x ) = \frac { \sqrt { 5.3 x ^ { 2 } + 50 x + 14 } } { 0.1 x + 0.8 }N(x)=0.1x+0.85.3x2+50x+14​​

An Efficiency Consultant Determines That When New Workers Are Hired $N ( x ) = \frac { \sqrt { 5.3 x ^ { 2 } + 50 x + 14 } } { 0.1 x + 0.8 }$