Asked by Yumin
Find limit if exists
1. Lim x goes to 0 x^2cos(1/x^3)
2.lim x goes to 1 to the left
4x/x^2+2x-3
3. Lim x goes to π to the right
cscx
Answers
Answered by
Steve
#1 cos(1/x^3) oscillates between 1 and -1, but the x^2 out front forces the limit to zero
http://www.wolframalpha.com/input/?i=x^2cos%281%2Fx^3%29+
#2 for x < 1, x^2+2x-3 < 0
So, you have 4*1/(small negative) -> -∞
#3 same thing here. sin(x) > 0 for x<π, so csc(x) -> +∞
You can type in your other functions at wolframalpha to verify the limits.
http://www.wolframalpha.com/input/?i=x^2cos%281%2Fx^3%29+
#2 for x < 1, x^2+2x-3 < 0
So, you have 4*1/(small negative) -> -∞
#3 same thing here. sin(x) > 0 for x<π, so csc(x) -> +∞
You can type in your other functions at wolframalpha to verify the limits.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.