Question

L'Hopitale's Rule: lim x-->0 of sinx^(cscx)

Answer 1

lny=(ln(sinx))(cscx)

Answer 2

proceed further using the product rule

Answer 3

couldn't i also change it to ln(sinx) / (sinx) ??

Answer 4

I did that and then took the derivative, but got another indeterminate.

Answer 5

1/0

Answer 6

what happens to sin(x) when x approaches 0?

Answer 7

it is 0

Answer 8

yes

what happens to csc(x) when x approaches 0?

Answer 9

be careful here: you'll need to approach 0 from both the left and right side

Answer 10

it might help to look at a graph or table

Answer 11

0

Answer 12

wait, it's indeterminate.

Answer 13

the limit isn't defined because the left hand limit and right hand limit are different

agreed?

Answer 14

right

Answer 15

so overall, the limit doesn't exist or it's not defined

Answer 16

so 0^infinity is the indeterminate?

Answer 17

well that's the thing, saying

\[\Large \lim_{x\to 0}\left[\csc(x)\right] = \infty\]

is a false statement

the limit does not exist

Answer 18

so what is the indeterminate?

Answer 19

it would have to be in these forms

https://upload.wikimedia.org/math/f/0/1/f01dca26ea8fa3e4af3440e5c6a629b3.png

Answer 20

if a piece of the limit does not exist, then the whole limit does not exist

Answer 21

so is the answer simply DNE?

Answer 22

Would it make a difference if it were x--> 0+

Answer 23

I guess to be more technical, I'm using this property

\[\LARGE \lim_{x \to a}\left\{\left[f(x)\right]^{g(x)}\right\} = {\lim_{x \to a}\left[f(x)\right]}^{\lim_{x \to a}\left[g(x)\right]}\]

Answer 24

In this case, a = 0, f(x) = sin(x) and g(x) = csc(x)

so,

\[\LARGE \lim_{x \to 0}\left\{\left[\sin(x)\right]^{\csc(x)}\right\} = {\lim_{x \to 0}\left[\sin(x)\right]}^{\lim_{x \to 0}\left[\csc(x)\right]}\]

Answer 25

the first limit, of sin(x) is 0

but the limit in the exponent DNE

Answer 26

@Amenah8

`Would it make a difference if it were x--> 0+`

sorry I didn't see that question til just now

Answer 27

yes it makes a big difference because

\[\Large \lim_{x \to 0^{+}}\left[\csc(x)\right]=\infty\]

Answer 28

so,

\[\LARGE \lim_{x \to 0^{+}}\left\{\left[\sin(x)\right]^{\csc(x)}\right\} = {\lim_{x \to 0^{+}}\left[\sin(x)\right]}^{\lim_{x \to 0^{+}}\left[\csc(x)\right]}\]

\[\LARGE \lim_{x \to 0^{+}}\left\{\left[\sin(x)\right]^{\csc(x)}\right\} = 0^{\infty}\]

which is one of the indeterminate forms

Answer 29

actually no I apologize, it's not one of the forms. I mixed up the terms on accident

Answer 30

so now I say: lny = ln(sinx)(cscx)

Answer 31

so the answer is infinity

Answer 32

\[\Large 0^{\infty} = 0\]

Answer 33

oh. :)

Answer 34

Think of it as \[\Large 0^x\] and let x approach infinity

well you can simplify 0^x to get 0

Answer 35

oh! i get it! thank you so much!!

Answer 36

no problem