Question

lim (sin x)^tan x as x->0+
... is this even possible? I got up to:

ln y=((sin x) Ln (sin x))/(cos x)

and then I got stuck... is there a different way to do this??

Answer 1

it would be lny= ln(sin x)sin x/cos x right?

Answer 2

now I think this would approach infinity as x approaches 0

Answer 3

I moved the (tan x) out in front bc of the exponential log rule, and then i changed tan x into (sin x / cos x)

Answer 4

that works I see we wrote the same thing just different ways! sorry.

Answer 5

Yeah i realized that after haha sorry about that! but you can't take the Ln (sin x) bc sin x is zero and that doesnt exist right? this is where i'm stuck, bc it 0*Ln sin 0 on the top, so do you just assume zero? and then do 0/1 and say the lim is 0? is it not possible bc of the Ln 0?

Answer 6

Assume the limit exists:
\[\lim_{x\to0^+}\sin^{\tan x}{x}=L\]
Then
\[\ln\left(\lim_{x\to0^+}\sin^{\tan x}{x}\right)=\ln L\]
The natural log function is continuous at positive non-zero arguments, so you have
\[\lim_{x\to0^+}\left(\ln\sin^{\tan x}{x}\right)=\ln L\]
Some log properties:
\[\lim_{x\to0^+}\tan x\ln\sin x=\ln L\]
The left hand side yields the indeterminate form \(0\cdot-\infty\), so rewriting that you have
\[\lim_{x\to0^+}\frac{\ln\sin x}{\cot x}=\ln L\]
Apply L'Hopital's rule:
\[\lim_{x\to0^+}\frac{\frac{\cos x}{\sin x}}{-\csc^2x}=\ln L\\
\lim_{x\to0^+}\frac{\cos x}{\sin x}\cdot(-\sin^2x)=\ln L\\
-\lim_{x\to0^+}\cos x\sin x=\ln L\\
0=\ln L\\
L=e^0=1
\]

Answer 7

ahh so you have to rewrite with cotangent! thank you so much!!