Question

5.
Find the limit of the function by using direct substitution. (6 points)
limit as x approaches zero of quantity x squared minus three.


3

Does not exist

-3

0
6.
Find the limit of the function by using direct substitution. (6 points)
limit as x approaches three of quantity x squared plus three x minus one.


17

0

-17

Does not exist
7.
Find the limit of the function algebraically. (6 points)
limit as x approaches four of quantity x squared minus sixteen divided by quantity x minus four.


Does not exist

4

1

8
8.
Find the limit of the function algebraically. (6 points)
limit as x approaches zero of quantity x squared minus two x divided by x to the fourth power.


Does not exist

8

0

-8

Answer 1

Finding limits by direct substitution means simply means to evaluate the function at the desired value: in the first case, we have to evaluate $f(x)=x^2-3$ at $x=0$: we have

$f(0)=0^2-3 = 0-3=-3$

Similarly, in the second example, we have

$f(x)=x^2+3x-1 \implies f(3) = 3^2+3\cdot 3-1 = 9+9-1 = 17$

Going on, we have

$f(x) = \dfrac{x^2-16}{x-4} = \dfrac{(x+4)(x-4)}{x-4} = x+4$

And thus we have

$f(4) = 4+4=8$

Finally, we have

$f(x) = \dfrac{x^2-2x}{x^4} = \dfrac{x(x-2)}{x^4} = \dfrac{x-2}{x^3}$

So, we can't evaluate this function at 0.

Answer 2

Answer:

previous was correct

Step-by-step explanation: