Question

what is the limit x-->2 of (x+1)^-1

Answer 1

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Answer 2

\[\Large\bf\sf \lim_{x\to2}\frac{1}{x+1}\]With limits, your first approach should be to plug the value directly into the limit.
If that causes a problem, we take a different approach.
We don't have any issues here though, plug in x=2 for your solution.

Answer 3

so the answer is simply 1/3?

Answer 4

yes

Answer 5

that just seems too obvious, the prof also wanted me to show where I used limit properties, any ideas for that?

Answer 6

hmm

Answer 7

We don't have a product, sum, difference, quotient of limits...
and no coefficient to pull out of the limit.
No properties to apply here.

We really just don't have anything special that we can do with this one except plug x=2 in. :\

Answer 8

ok thanks for your time!

Answer 9

np

Answer 10

got time for 1 more?

Answer 11

sure :3

Answer 12

let f(x) = x+1/ 4x-2 determine the point of discontinuity

Answer 13

I figured it would be 1/2 right?

Answer 14

Yes.
It's not too bad with this function.
When they get a little harder, it's important to understand the process though.

You set the denominator = 0.
Then solve for x, that will give you points of discontinuity.

4x-2 = 0

Answer 15

ohh and x can also not be -1 because the numerator cannot be zero either

Answer 16

Think back to your math rules a sec.
The golden rule is `you cannot divide by 0`.
That means zero in the denominator is a no no!
Very bad.

What about 0/5?
Maybe put it into your calculator if you're unsure.

Answer 17

If you take zero and divide it into 5 pieces, you still have zero.
0/5 = 0.

The numerator `can` equal zero. Make sure you understand the difference! :D

Answer 18

ok so then 1/2 was the only point of discontinuity, now heres the second half of the question,
can this point of disc be removed to make continuous for all x?

Answer 19

I'm probably over thinking this.
I guess they're just asking, if we write the problem like,\[\Large\bf\sf f(x)=\frac{x+1}{4x-2},\qquad x\ne1/2\]Putting the restriction on x,
Removing the discontinuity,

then yes I guess the function would be continuous over it's entire domain.

Answer 20

ok I think I got it thnks