# Inverse func Limits

1. Apr 17, 2007

### f(x)

1. The problem statement, all variables and given/known data
Evaluate
$$\lim_{x\rightarrow 0} \ \frac{sin(tanx)-tan(sinx)}{sin^{-1}(tan^{-1}x)-tan^{-1}(sin^{-1}x)}$$

2. Relevant equations

3. 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

2. Apr 17, 2007

### Gib Z

how many times have you tried the rule?

3. Apr 17, 2007

### f(x)

I applied L^Hopital's once.

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

4. Apr 17, 2007

### Gib Z

Yup I know what you meant with the arctan thing.

Try using l'hopitals rule again.

5. Apr 17, 2007

### 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

6. Apr 17, 2007

### 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 ?

7. Apr 17, 2007

### HallsofIvy

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

8. Apr 17, 2007

### 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.

9. Apr 18, 2007

### f(x)

Just forgot to put the square...thx for bringing it into notice.
I havent seen limits with inverse trig functions in them . And now i cant even try substitutions as Halls said . If any1 figures a way out, do hint it here

10. Apr 18, 2007

### Gib Z

I already told you! Use L'hopitals again! It wont be pretty but bad luck.

11. Apr 22, 2007

### f(x)

Finally found the solution -:

12. Apr 22, 2007

### 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.

13. Apr 22, 2007

### matt grime

L'Hopital is trucating Taylor series.

14. Apr 22, 2007

### Gib Z

Could you provide more info on that, I dont quite follow...In the link the solver truncated after 4 terms, so lhopital wud work after 4 tries?

15. Apr 22, 2007

### 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?

16. Apr 22, 2007

### f(x)

I dont understand whats o(x^7) mean ? Could some1 explain

17. Apr 22, 2007

### 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.