Question

Solve using L'Hôpital's rule:

Lim x goes to 0 from the right , (cos x)^1/x

Answer 1

\[\lim_{x \rightarrow 0+} (\cos x)^{1/x}\]

Answer 2

\[Ln ((\cos x)^{\frac{ 1 }{ x }})=\frac{ 1 }{ x }\ln(\cos x)\]
Apply l'Hopital's rule on f(x)=ln(cos x) and g(x)=x to find the limit of the function above. Then use exponential to find your final answer.

Answer 3

Thanks , so :

if we tried the limit we'll get 0/0 and now we could use the rule , so:

ln cosx/x = 1/cosx/x= 0

?

Answer 4

No, you calculate \[\lim_{x \rightarrow 0^{+}}\frac{ f'(x) }{ g'(x) }\] L'Hôpital's rule stipulates that \[\lim_{x \rightarrow 0^{+}}\frac{ f'(x) }{ g'(x) }=\lim_{x \rightarrow 0^{+}}\frac{ f(x) }{ g(x) }=L\] When you find that limit, exp(L) will be your final answer.

Answer 5

That's what I did , 1*ln(cosx)/x =

ln (cosx)/x=

the derivative of ( ln(cos x)) = 1/cox
the derivative of x =1
so 1/cosx/1 = 1/1=1 ?

Answer 6

You have a problem on the derivative of ln(cos x)).
\[(\ln (u))'=\frac{ u' }{ u }\]

Answer 7

Thank you for being patient with me ^^

I used my notes and it says the derivative of ln x = 1/x

anyways we'll :
-sinx/cosx/1
so 0/1/1 = 0 ?

Answer 8

That's correct :)
Now what would be the final answer?

Answer 9

ln(0)=1

Answer 10

Yes, Gz!

Answer 11

Thank you ^^

Answer 12

You're welcome ^^