How Does L'Hopital's Rule Apply to Lim (1/x)^tan(x) as x Approaches 0?

[felipe oteiza]

l'hopital must be apply, i'll be very grateful

 

[BvU]

Hello felipe

Do you know the limit for $\sin x \over x$ ?

 

[BvU]

oops, sorry, you mean $$x^{-\tan x}\ \ ?$$

 

[felipe oteiza]

oops, sorry, you mean $$x^{-\tan x}\ \ ?$$

yes! the last

 

[BvU]

Where does $\ x^{\tan x}\$ go for $\ x\downarrow 0$ ?

 

[felipe oteiza]

Where does $\ x^{\tan x}\$ go for $\ x\downarrow 0$ ?

lim ( 1/x )^tan x as x->0

 

[BvU]

Yes, that was my question

 

[felipe oteiza]

Yes, that was my question

I don't understand your question (my english is not very good)

 

[BvU]

What is the limit $\ \ \displaystyle \lim_{x\downarrow 0} \ x^{\tan x}\$ ?

 

[mathman]

tanx ~ x as x ->0, so problem can be looked at as $\lim_{x-->0} x^x$ However $x^x=e^{xlnx}$.
Since $\lim_{x->0}xlnx=0$, the final answer = 1.

 

[felipe oteiza]

tanx ~ x as x ->0, so problem can be looked at as $\lim_{x-->0} x^x$ However $x^x=e^{xlnx}$.
Since $\lim_{x->0}xlnx=0$, the final answer = 1.

thanks

 

[shaztp]

Tan (0)=0 there for answer will be 1

 

[mathman]

Tan (0)=0 there for answer will be 1

Not by itself. The function is $(\frac{1}{x})^{tanx}$, so as x->0, the expression becomes $(\frac{1}{0})^{0}$ which is indeterminate.