anonymous
  • anonymous
Help with this problem plz!!
Mathematics
schrodinger
  • schrodinger
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
1) lim x->0 + f(x) 2) lim x->0 f(x) 3) lim lim x->1 - f(x)
1 Attachment
anonymous
  • anonymous
4) lim x-> -1 f(x) 5) lim x-> 2- f(x) 6) lim x-> 2+ f(x) 7) lim x-> 2 f(x) 8) lim x-> -2 f(x)
anonymous
  • anonymous
9) lim x-> 3f(x)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@ParthKohli can u help?
AravindG
  • AravindG
oh 8 questions??
AravindG
  • AravindG
which of these you have doubt on?
anonymous
  • anonymous
yea but u dont have to help me with all of them
anonymous
  • anonymous
can we start with the first one?
AravindG
  • AravindG
OK :)
anonymous
  • anonymous
did he answer?
anonymous
  • anonymous
no
anonymous
  • anonymous
1. the limit is 2 approaching for the right 2. the limit dne because it has two limits from both sides and is not continuous. 3. the limit is 1
anonymous
  • anonymous
okay thanks i'll do the rest
anonymous
  • anonymous
http://calculus.nipissingu.ca/problems/limit_problems.html here's a website that could help you. good luck

Looking for something else?

Not the answer you are looking for? Search for more explanations.