Asked by Shaniquaa
limit as x-> 0
(1+sinx)^(cscx)
How do I show that this is indeterminate? Can I say that 1^(1/0) = 1^infinity?
(1+sinx)^(cscx)
How do I show that this is indeterminate? Can I say that 1^(1/0) = 1^infinity?
Answers
Answered by
Reiny
have you seen the definition of e ?
e = lim(1 + 1/n)^n as n --> ∞
or e = lim(1 + n)^(1/n) as n ---> 0
now look at your limit, does it not resemble my last statement?
lim(1+sinx)^(cscx) as x ---> 0
= lim(1+sinx)^(1/sinx) as x ---> 0
= e = 2.7182818...
you can test this on a calculator
set it to radians, and let x = .000001
I get 2.71828147, pretty close to e
e = lim(1 + 1/n)^n as n --> ∞
or e = lim(1 + n)^(1/n) as n ---> 0
now look at your limit, does it not resemble my last statement?
lim(1+sinx)^(cscx) as x ---> 0
= lim(1+sinx)^(1/sinx) as x ---> 0
= e = 2.7182818...
you can test this on a calculator
set it to radians, and let x = .000001
I get 2.71828147, pretty close to e
Answered by
Shaniquaa
I don't need the limit, I just want to know how to show it is in the indeterminate form.
Answered by
Reiny
But..
as I showed in my solution, it isn't indeterminate,
it works out to the constant e, or 2.7....
as I showed in my solution, it isn't indeterminate,
it works out to the constant e, or 2.7....
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.