Question

I could use some help to solve this problem.

Lim X^(tanx)
X-->0

I am fairly sure I would need to use l'hospitals rule but not sure how to convert the x^tanx into a fraction

Answer 1

yes, this one is tricky.
one way is write x = \( e^{\ln x}\)

Answer 2

so your problem is a limit for
\[ x^{\tan x} = \left( e^{\ln x}\right)^{\tan x}= e^{\ln x \cdot \tan x} \]

Answer 3

is there a rule that will show me why you did what you did?

Answer 4

now you do
\[ \lim_{x \rightarrow 0} \ln x \tan x \]

Answer 5

Other way of presenting phi's trick: \(x^{\tan x} = e^{\ln(x^{\tan x})} = e^{\tan x \ln x}\).
Same result.

Answer 6

by definition

\[ e^{\ln x }= x \]

Answer 7

Lol I honestly did not know that

Answer 8

meanwhile, the expression is now a product, so we still have to massage it into a quotient

Answer 9

So I take apply l'hospitals to

lim lnxtanx
X-->0

Answer 10

So

Lnx/tanx^-1
?

Answer 11

I haven't worked it out. It will take some fiddling to get this.

Answer 12

Ok, quick question

What can I search in Google to teach me how to solve this?

Converting x^tanx into a fraction

Answer 13

Or is there a rule I need to know? Because product rule wouldn't work.

Answer 14

you can only try.
I would not give up (yet) on what we are doing

Answer 15

Ok will return

Answer 16

Did you apply L'Hospital rule already?

Answer 17

Lol I did this wrong entirely

I applied the product rule to the exponent

I got

e^(lnxsec^2x + tanx/x)

Answer 18

This one is very confusing for me, I will need to find some YouTube videos

Answer 19

on \(\frac{\ln x}{\frac{\cos x}{\sin x}}\), I obtain,
\[
\lim_{x\to 0} \frac{1/x}{-1/\sin^2x} = \lim_{x\to0} -\frac{\sin^2x}{x}
\]
If you know that \(\frac{\sin x}{x}\to 1\) as \(x\to 0\), then you're done. Otherwise, apply L'Hospital rule again.

Answer 20

yes, that looks good.

Answer 21

so the original limit is
e^0 = 1

Answer 22

So essentially what you are doing is changing x^tanx

To

lnX/tanx

And tanx = cosx/sinx
?

Answer 23

And then going from there

LnX
=
1/x

Answer 24

Is that an identity going from
Cosx/sinx to -1/sin^2x

Answer 25

that is the derivative

Answer 26

I lol That just didn't click that we went from lnX to 1/x which was done by taking the derivative. So it would make sense to take derivative of the bottom

Answer 27

\(\ln x\tan x = \frac{\ln x}{\frac{\cos x}{\sin x}}\)

Answer 28

if we summarize, we get to

ln x tan x = ln x / cot x
this is now in the form oo/oo
take the derivatives: 1/x and -1/sin^x
and rewrite that as - sin^2(x )/ x

if you don't the the sin x / x limit, then apply again:
-2 sin x cos x / 1
and the limit is -2*0*1= 0

and that 0 is the limit of
\[ e^{\lim \ln x \tan x} = e^0 = 1\]

Answer 29

Ok thank you, once closed can I come back to this question?

Answer 30

Like to look over?

Answer 31

I think it's possible. Also, you can find it easily on your profile page.