Question

tricky or nah?
\(\huge \lim_{x \rightarrow 0^+} (\sin x)^{\frac{1}{\ln x}} \)

Answer 1

Not at all.

Answer 2

:)

Answer 3

Did you mean this
\[\huge \lim_{x \rightarrow 0^+} \sin (x^{\frac{1}{\ln x}}) \]

or this
\[\huge \lim_{x \rightarrow 0^+} (\sin x)^{\frac{1}{\ln x}} \]

Answer 4

second one hahaha
this is for the people who's just taken calc1 not the god level like you, kai

Answer 5

I wonder if there is a clever way to work this with out using Lhopital

Answer 6

without you mean? I do not know without using L'Hospital, but I know using L'Hospital

Answer 7

If I may use the small angle approximation, \(\sin x\approx x\), and prentend to ignore all technicalities of singularities :
\[x^{1/\ln x} = e^{\ln x/\ln x} = e\]

Answer 8

oooh that's interesting

Answer 9

That gives us some number but it raises many serious issues, we won't know if that number is correct or not w/o addressing them...

Answer 10

here's what I thought: start out by
\(\large y(x) = (sin~x)^{1/ln~x} \rightarrow ln~y = (sin~x)^{1/ln~x} \rightarrow \color{blue}{ln~y = \frac{ln~(sin~x)}{ln~x}}\)

then work the limit of \(ln~y \) giving indeterminate form -inf/-inf

Answer 11

gives us e

Answer 12

that is a less unorthodox... and nobody should have any issues with that :)

Answer 13

typo
\(\large y(x) = (sin~x)^{1/ln~x} \rightarrow ln~y = ln (sin~x)^{1/ln~x} \rightarrow \color{blue}{ln~y = \frac{ln~(sin~x)}{ln~x}} \)

Answer 14

we may directly use this property of continuous functions :

\[\lim f(g(x)) = f(\lim g(x))\]

Answer 15

\[\lim e^{g(x)} = e^{\lim g(x)}\]

Answer 16

\[\lim (\sin x)^{1/\ln x}\\~\\
=\lim e^{\ln (\sin x)^{1/\ln x}}\\~\\
=\lim e^{\ln (\sin x)/\ln x}\\~\\
= e^{\lim \ln(\sin x)/\ln x}
\]

Answer 17

thats not much different than what you're doing, but it certanly looks better because we're not introducing new variables and it allows you to go in single direction...

Answer 18

I want to medal you again

Answer 19

Haha all that is possible because of this
\[x = e^{\ln x}\]

Answer 20

assuming x > 0

Answer 21

Continuing where ganeshie8 left off.

\(\color{#000000 }{ \displaystyle {\lim_{x \to 0}~\ln(\sin (x))/\ln(x)} }\)

(-∞)/(-∞), so L'H'S is applied.

\(\color{#000000 }{ \displaystyle {\lim_{x \to 0}~\frac{\frac{\cos(x)}{\sin(x)}}{\frac{1}{x}}} }\)

0/0, so L'H'S is applied again

\(\color{#000000 }{ \displaystyle {\lim_{x \to 0}~\frac{\frac{-1}{\sin^2(x)}}{\frac{1}{-x^2}}}={\lim_{x \to 0}~\frac{x^2}{\sin^2x} }=\left[{\lim_{x \to 0}~\frac{x}{\sin x} }\right]^2 }\)

\(\color{#000000 }{ \displaystyle }\)

Answer 22

and e^0 ... blah blah blah

Answer 23

lol why