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?

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
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....
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