lim(as p-->4) of (p^3 - 64) / (4-p)
My work:
p^3 - 64 = (p-4)(p^2 + 4p + 16)
4 - p = -(p-4)
((p-4)(p^2 + 4p + 16)) / -(p-4)
p-4 cancels out
-(p^2 + 4p + 16)
-((4)^2 + 4(4) + 16)
-(16 + 16 + 16)
- 48
Is this right?
looks good to me
Yes, your work is correct. Let's go over the process step by step to verify the result.
We start with the expression:
(p^3 - 64) / (4 - p)
Next, we factorize the numerator using the difference of cubes formula:
p^3 - 64 = (p - 4)(p^2 + 4p + 16)
The denominator, 4 - p, can be rewritten as -(p - 4) since it is the opposite sign of (p - 4).
Now, we can cancel out the common factor of (p - 4) in the numerator and the denominator.
This leaves us with:
(p^2 + 4p + 16) / -(p - 4)
Now, we can take the limit as p approaches 4. Plugging in p = 4 into the expression, we get:
(-(4^2 + 4(4) + 16)) / -(4 - 4)
= -(16 + 16 + 16) / -0
= -48 / 0
However, dividing by zero is undefined, so the limit does not exist.
To summarize, you correctly factored the numerator and simplified the expression. However, the final result is not a specific value but rather undefined due to division by zero.