limit as x-> 0

(1+sinx)^(cscx)

How do I show that this is indeterminate? Can I say that 1^(1/0) = 1^infinity?

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

I don't need the limit, I just want to know how to show it is in the indeterminate form.

But..

as I showed in my solution, it isn't indeterminate,
it works out to the constant e, or 2.7....

To determine the limit as x approaches 0 of the function (1 + sin(x))^(csc(x)), we first need to understand if it is an indeterminate form.

To do this, we can rewrite the expression as e^(ln((1 + sin(x))^(csc(x))) and then evaluate the limit using properties of exponential and logarithmic functions.

Let's break down the steps:

Step 1: Rewrite the function using the natural logarithm.
(1 + sin(x))^(csc(x)) = e^(ln((1 + sin(x))^(csc(x))))

Step 2: Simplify the expression using logarithmic properties.
ln((1 + sin(x))^(csc(x))) = csc(x) * ln(1 + sin(x))

Step 3: Evaluate the limit.
Now, we can take the limit as x approaches 0. Let's break it down further.

Taking the limit of csc(x) as x approaches 0:
lim(csc(x)) = lim(1/sin(x))
Since sin(x) approaches 0 as x approaches 0, the limit of csc(x) is infinity.

Taking the limit of ln(1 + sin(x)) as x approaches 0:
lim(ln(1 + sin(x))) = ln(1 + sin(0))
Since sin(0) = 0, ln(1) = 0.

Step 4: Combine the limits.
lim((1 + sin(x))^(csc(x))) = e^(lim(csc(x) * ln(1 + sin(x))))
= e^(infinity * 0)
= e^indeterminate form

We can see that the limit of (1 + sin(x))^(csc(x)) is an indeterminate form since it is in the form of e^(indeterminate form).

So, to answer your first question, yes, the limit is indeterminate.

Regarding your second question, it is not correct to say that 1^(1/0) = 1^infinity. The expression 1^(1/0) is undefined because dividing by zero is undefined in mathematics.

In conclusion, the limit as x approaches 0 of (1 + sin(x))^(csc(x)) is an indeterminate form and cannot be simplified further without additional analysis.