Question

lim (sin x )^tanx as x->0 ?

Answer 1

I think you can apply l'hopitals rule, after modifying it a bit...

Answer 2

Yeah,just solved it with these rules,it's equal 1.

Answer 3

Yep :)

Answer 4

I had to try it myself... I'm guessing you ended up with something like \[\Large \lim_{x \rightarrow 0} e^{-\cos x \sin x}\]which if you plug in 0 gives you 1.

Answer 5

After modifying it using logs, then using l'hopitals rule.

Answer 6

Can you please write the whole way of it ? I did it a bit different ...

Answer 7

lim (sin x )^tanx as x->0

First let y = sinx^tanx, and take logs of both sides \[\large \ln y = \tan x * \ln (\sin x)\] put both sides to the power of e
\[\LARGE y = e^{\tan x *\ln (\sin x)}\]

Answer 8

So that was just to modify the limit. Then write it as \[\LARGE \lim_{x \rightarrow 0 } e^{\tan x *\ln (\sin x)}\] for which you can bring the limit up into the exponent \[\huge e^{ \left( \lim_{ x \rightarrow 0}\tan x *\ln (\sin x) \right)} \]

Answer 9

Site crashed... that's why i write solutions in pieces :P

To apply l'hopitals rule, get the tanx into the denominator so that it's in inf/inf form: \[\huge e^{ \left( \lim_{ x \rightarrow 0}\frac{ \ln (\sin x) }{ \tan^{-1} x } \right)}\] cos now if you plug in 0 you get -inf/inf

Answer 10

Now you can differentiate using hopitals
\[\huge e^{ \left( \lim_{ x \rightarrow 0}\frac{ \frac{ \cos x }{\sin x }}{ - \cot^2 x } \right)}\]
which after simplifying gives what i gave above.

Answer 11

Thank you very much ! :)

Answer 12

No prob, i'm just satisfied i could actually solve it! hehe

Answer 13

\[\tan^{-1}(0)=0\]

Answer 14

oh..i see what you did...you should not use that notation

Answer 15

ohh, yes. My bad.

Answer 16

But it's not too hard to figure out what i meant in this case, since i mentioned putting it into the denominator.

Answer 17

if you want 1 over the tangent you should write \[\frac{1}{\tan(x)}\] or

\[\left[\tan(x)\right]^{-1}\]



\[tan^{-1}(x)\] typically means the inverse tangent

Answer 18

haha, i know :P

Answer 19

I wrote it as (tanx)^-1 and then thought "no, tan^-1 x is neater..."

Answer 20

I would have known that if i actually read from the beginning and not just your last post ;)

Answer 21

haha :)

Answer 22

cheers!