Help me to verify the piecewise function answers

macjack
Messages
11
Reaction score
0
I am trying to solve the exercise in this link, please let me know if i am doing something wrong. ( i didn't find answers to this exercise that's why i am sending it here)

http://www.libraryofmath.com/limits-with-piecewise-functions.html


exercise 1
----------
a) lim x->3- f(x) = 7
b) lim x->3+ f(x) = 7
c) lim x->3 f(x) = 7 because both one-sided limits are same.

exercise 2
----------
a) lim x->0- f(x) = 2
b) lim x->0+ f(x) = 1
c) lim x->0 f(x) = no limit (because one-sided limits are not same)
d) lim x->2- f(x) = 1
e) lim x->2+ f(x) = 3
f) lim x->2 f(x) = no limit (because one-sided limits are not same)

exercise 3
----------
a) lim x->-1- f(x) = 1
b) lim x->-1+ f(x) = 1
c) lim x->-1 f(x) = 1
d) lim x->0- f(x) = 0
e) lim x->0+ f(x) = 0
f) lim x->0 f(x) = 0
g) lim x->1- f(x) = 1
h) lim x->1+ f(x) = 1
i) lim x->1 f(x) = 1

Thanks for your time.
 
Physics news on Phys.org
macjack said:
I am trying to solve the exercise in this link, please let me know if i am doing something wrong. ( i didn't find answers to this exercise that's why i am sending it here)

http://www.libraryofmath.com/limits-with-piecewise-functions.html


exercise 1
----------
a) lim x->3- f(x) = 7
b) lim x->3+ f(x) = 7
c) lim x->3 f(x) = 7 because both one-sided limits are same.

exercise 2
----------
a) lim x->0- f(x) = 2
b) lim x->0+ f(x) = 1
c) lim x->0 f(x) = no limit (because one-sided limits are not same)
d) lim x->2- f(x) = 1
e) lim x->2+ f(x) = 3
f) lim x->2 f(x) = no limit (because one-sided limits are not same)

exercise 3
----------
a) lim x->-1- f(x) = 1
b) lim x->-1+ f(x) = 1
c) lim x->-1 f(x) = 1
d) lim x->0- f(x) = 0
e) lim x->0+ f(x) = 0
f) lim x->0 f(x) = 0
g) lim x->1- f(x) = 1
h) lim x->1+ f(x) = 1
i) lim x->1 f(x) = 1

Thanks for your time.

Yes, looks good to me. Congratulations. :wink:

Btw, you haven't done the last 2 problems. :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top