Help with problem below. I'm finding it difficult, but I did come up with 3. Can anyone check my solution?

Question

. . = x^2 + 3, if x < 0
. . = 0, if x = 0
f(x)= x^2 - 3 if 0 < x < 2
. . = 1 if x = 2
. . = x^3 - 7 if 2 < x.


Then lim . . . . . . . f(x) =
. . . . . . . . x=>2

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Possible solutions from the book...

a. -3

b. -1

c. 1

d. 3

e. none of these

Answer 1

It is e....... none of these because:
1)If it was a. then, (-3)^3 -7
which is -27-3 and is coming 30 which is not possible for the sum.
2)If it was b. then, (-1)^3-7
which is -1-3 and is coming -4 which is not possible for the sum.
3)If it was c. then, (1)^3-7
which is 1-7 and the answer for this is -6. (Again not possible)
4)If it was d. then, the sum would be
27-7
which = 20......so therefore it is E.

Answer 2

It's been a while since I've done limits so maybe I am completely off, but I come up with e) none of these. You already know that when x=2, f(x)=1, which means that when x=2, f(x) is the same as when 2 < x... plug 2 for x into x^3 - 7 and you get 1 (2^3=8 8-7=1), so you know that whatever f(x) is, it's slope is changing uniformly at x=2 and at x > 2. Plus you are already given that f(x)=x^3 - 7 when 2 < x, which is the same as saying x > 2. So, as x gets bigger than 2 in the function x^3 - 7, f(x) is going to go off into infinity. Hence the limit is oo (infinity).

Sorry cute snowball... we're not talking about simply plugging in a standard value for x, but rather finding the limit of f(x) as x = > 2.

Answer 3

At 2 the limit X=>2-(from the left ) is 1 (2^2-3)
The limit X=>2+ is also 1 (2^3-7)
So the limit exists and is 1 .Answer c)

Answer 4

c