# Limit of arc tan.

1. Jan 29, 2007

### SheldonG

1. The problem statement, all variables and given/known data
Evalute:

a) lim (x->0) (arctan x)/x
b) lim (x->1) (arctan(x) - pi/4)/(x-1)

2. Relevant equations
Inverse tangent, trig identities. Kline's calculus, which I am teaching myself from, does not have that much detail on limits.

3. The attempt at a solution

For (a), I solved it fairly easily. I used y = arctan(x), x = tan(y) to rewrite the limit as

y/tan(y) = y/(sin y/cos y) = y(cosy)/sin y) = (y/sin y)(cos y).

as x->0, y/sin y = 1, as does cos(y).

Applying the same approach to (b) has not rewarded me with a solution.

Making the same substition y = arctan(x), x=tan(y), I get

(y - pi/4)/(tan(y) - 1)

(4y - pi)(cos y)/(4(sin y - cos y))

But I don't seem to be able to find my way to a solution.

Some things I have tried:

trig identities like -cos(2a) = sin^2 y - cos^2 y, as well as others.

I tried dividing up the fraction, and computing the sums:

y/(sin y - cos y) - pi/(sin y - cos y).

I also tried finding the limit of the difference between this and part (a), on the assumption that the limit of the sums is the sum of the limits.

If anyone could prod this old guy in a useful direction, it would be appreciated.

Thanks,
Sheldon

2. Jan 29, 2007

### mjsd

when x->1, top and bottom goes to zero, can use l'hopital rule

3. Jan 29, 2007

### SheldonG

Thanks mjsd, but the book hasn't covered that yet.

4. Jan 29, 2007

### Gib Z

Well then this is a good time to learn. L'hopitals Rule is very useful to apply to limits. Basically, if by direction substitution you get 0/0, infinity/infinity or some other indeterminate forms, then you can use this method.

Say your limit is x>C f(x)/g(x), where C is some number. If we get an inderterminate form, the limit is equal to x>C f'(x)/g'(x). We just took the derivative of the top and bottom. Then we try substituting again. That is what out original limit is equal to. If it doesn't work, and we get another indeterminatite form, use the rule again and again til it works.

5. Jan 29, 2007

### SheldonG

Hi Gib Z.

Perhaps this would be a good time to learn... But working on the assumption that it is solvable without L'hopitals Rule, I wonder if anyone has any other suggestions?

Sheldon

6. Jan 29, 2007

### IMDerek

Maybe using the power series for arctan (x) would be useful here.
arctan x = x - x^3/3 + x^5/5 - ...

Part b is just the definition of derivative for arctan (x) at x=1.

d/dx (arctan x) = (1+x^2)^-1

EDIT: part a is also the definition of derivative for arctan(x) at x=0. It should be more clear if you write it as [arctan(x)-0]/(x-0).

Last edited: Jan 30, 2007
7. Jan 30, 2007

### SheldonG

Ah, thank you, Derek. I should have seen that. That makes sense. So I can say that

(arctan(x) - pi/4)/(x-1) = (arctan(x) - arctan(pi/4))/x - 1

so it would follow right away that the limit is the same as the derivative of (b), which gives me 1/2. Wonderful. Thank you so much.

Sheldon