Evaluate Inverse Func Limits: Solving 0/0 Form

[f(x)]

Homework Statement

Evaluate
$$\lim_{x\rightarrow 0} \ \frac{sin(tanx)-tan(sinx)}{sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x)}$$

Homework Equations

The Attempt at a Solution

I have no idea about evaluating inverse function limits.(denominator)
I tried L^Hopital's rule but it the derivatives still remain in 0/0 form
Thx for any help

 

[Gib Z]

how many times have you tried the rule?

 

[f(x)]

how many times have you tried the rule?

I applied L^Hopital's once.

BTW, by $\ tan^{-1}x \$ I mean $\ \ arctanx , not \frac{1}{tanx}$

 

[Gib Z]

Yup I know what you meant with the arctan thing.

Try using l'hospital's rule again.

 

[f(x)]

Here's what i have,

$$D^r \ = \ sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x)$$

$$D^r'\ = \ \frac{1}{ \sqrt{1-tan^{-1}x}}. \frac{1}{1+x^2} - \frac{1}{(1+sin^{-1}x)^2}. \frac {1}{ \sqrt {1-x^2} }$$

Now i use this- :
As $\ x \ \rightarrow 0 , tan^{-1}x \ \rightarrow 0$
As $\ x \ \rightarrow 0 , sin^{-1}x \ \rightarrow 0$

So, my denominator goes 0.
Taking another derivative would be nasty

 

[f(x)]

Whoa...i just got an idea.
Could i try replacing $\ sin^{-1}x$ by $\ tanx$ and so on as all tend to 0 with x tending to 0 ?

 

[HallsofIvy]

It doesn't follow that $sin^{-1} x$ and $tan x$ tend to 0 at the same rate and that is the crucial thing.

 

[Gib Z]

You didn't differentiate properly.

The derivative for the bottom should be
$$\frac{1}{1+x^2} \frac{1}{\sqrt{1-(\tan^{-1}x)^2}} - \frac{1}{\sqrt{1-x^2}} \frac{1}{1+(\sin^{-1}x)^2}$$.

Either way, the denominator still goes to zero. Sure its nasty, but just do it. If you really can't be bothered, type it into here: www.calc101.com. It'll find the derivative for you, but entering it is just as much of a hassle as find the derivative yourself.

 

[f(x)]

You didn't differentiate properly.

Just forgot to put the square...thx for bringing it into notice.
I haven't seen limits with inverse trig functions in them . And now i can't even try substitutions as Halls said . If any1 figures a way out, do hint it here

 

[Gib Z]

I already told you! Use l'hospital's again! It won't be pretty but bad luck.

 

[f(x)]

Finally found the solution -:
LINK

 

[Gib Z]

Thats quite a nice solution. Really he just truncated the taylor series and computed it that way, so its not such a new method, but none of us thought of that so good work.

 

[matt grime]

L'Hopital is trucating Taylor series.

 

[Gib Z]

Could you provide more info on that, I don't quite follow...In the link the solver truncated after 4 terms, so lhopital wud work after 4 tries?

 

[matt grime]

If you want the limit of f(x)/g(x) as x tends to zero (say). Take the Taylor series of each about 0. You can read off the limit. That is what L'Hopital's rule is: looking at the leading coeffecients of Taylor series. What else do you think you're doing when differentiating and plugging in other than finding the coefficients of a Taylor series?

 

[f(x)]

http://www.mathlinks.ro/Forum/latexrender/pictures/4/d/8/4d8d10bdf658609e9d8a17b2660fd35707726e8a.gif
I don't understand what's o(x^7) mean ? Could some1 explain

 

[Gib Z]

Its sort of an error term. Really the series for sin and tan have an infinite number of terms, but he just just terms up to x^7, then o(x^7), or terms less important that x^7. When he says less important than, he means the error is less than. So if x is 1, the approximation we got from the first 4 can not be more than 1 off. This is what I THINK, learning from Hurkyl's posts in my thread in Calc (Not HW section).

Thanks Matt for the reply, i understand now.