Given p(x) = x^2 - 1, q(x) = x^3 - 1, and r(x) = x^4 - 1, evaluate:?

Question

lim p(x) .
x=>1 q(x)

What rule did you use?

it is the L'Hopital's rule

sorry about that

lim
x=>oo

p(x) / q(x)

Answer 1

I don't think I've seen a limit problem set up this way.
lim p(x). ?
I'll give two possible answers. Is this exactly the way the problem is written? Very odd a question would be written this way. I've taught and researched math for five years and have never seen a problem stated in this format. Here are two possible answers to this. Hope I can help out!
If your trying the find the limit q(x)->1 the it's easy. q(x) = x^3 - 1
Thus, (1)^3 - 1 = 0
Since we have the limit of q(x) = 0.
If the "p(x)" was not written in here the answer would be zero and you would be done. The p(x) makes me suspicious!

Now, take that answer of the limit you found to find the limit of p(x). So then we have p(x)->0
p(x) = x^2 - 1 So, (0)^2 - 1 = -1. So your final answer would be
-1.
Only possible answer if this is the true format.
What rule? No rules are really needed here. Just basically plugging in a number and receiving the put out. We could factor out some of these equations but the answer would still be the same. If there was division, q(x)/p(x), then we could take derivatives using L'Hopital's rule. But I don't see any sign of division here.

Answer 2

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Answer 3

as x =>1, the limit of all functions above is 0

Answer 4

The answer is 0 (zero)

Answer 5

you are either really smart or really funny...

Answer 6

wtf