Arctan limit (without L'Hopital's Rule)

[AlexandraMarie112]

Homework Statement

Limx-->positive infty arctan(1+x)/(1-x)

Homework Equations

The Attempt at a Solution

I just need to know if my answer is right.

Knowing that when the leading coefficients of the x when its the same, then the answer is just the ratio. So it would be -1. Then in my calculator I put arctan -1 and it gave -45. Is this right?

 

[Mark44]

Homework Statement

Limx-->positive infty arctan(1+x)/(1-x)

Homework Equations

The Attempt at a Solution

I just need to know if my answer is right.

Knowing that when the leading coefficients of the x when its the same, then the answer is just the ratio. So it would be -1. Then in my calculator I put arctan -1 and it gave -45. Is this right?

In math problems, trig functions and their inverses are usually in radians. Your calculator is set to degree mode. The book's answer will almost certainly be in radians. Otherwise, your value of -1 is correct.

 

[AlexandraMarie112]

In math problems, trig functions and their inverses are usually in radians. Your calculator is set to degree mode. The book's answer will almost certainly be in radians. Otherwise, your value of -1 is correct.

The answer is not in the book. My book has a weird way of giving the answers. Only gives for odd numbered questions and this ones an even so I have no way of knowing if my answer is right.

So instead of -45 it would be -pi/4 ?

 

[Ray Vickson]

Homework Statement

Limx-->positive infty arctan(1+x)/(1-x)

Homework Equations

The Attempt at a Solution

I just need to know if my answer is right.

Knowing that when the leading coefficients of the x when its the same, then the answer is just the ratio. So it would be -1. Then in my calculator I put arctan -1 and it gave -45. Is this right?

Do you mean
$$\frac{\arctan(1+x)}{1-x}$$
or do you mean
$$\arctan \left( \frac{1+x}{1-x} \right)?$$
If you mean the latter you need parentheses, like this: arctan((1+x)/(1-x)) or arctan[(1-x)/(1+x)].

 

[Mark44]

The answer is not in the book. My book has a weird way of giving the answers. Only gives for odd numbered questions and this ones an even so I have no way of knowing if my answer is right.

That's not at all unusual for math books. If the answer you get for an odd-numbered problem agrees with the book's answer, you can have some confidence that your work for a nearby even-numbered problem is also correct.

So instead of -45 it would be -pi/4 ?

Right.

If you mean the latter you need parentheses, like this: arctan((1+x)/(1-x)) or arctan[(1-x)/(1+x)].

I agree completely. I assumed from the answer given by the OP, that the problem was as you show it here, Ray.

 

[Ray Vickson]

That's not at all unusual for math books. If the answer you get for an odd-numbered problem agrees with the book's answer, you can have some confidence that your work for a nearby even-numbered problem is also correct.

Right.

I agree completely. I assumed from the answer given by the OP, that the problem was as you show it here, Ray.

I agree that we can try to guess the meaning by looking at what the OP has done; but the OP needs to realize that clearer communication is important.

 

[Mark44]

but the OP needs to realize that clearer communication is important.

Absolutely.

 

[AlexandraMarie112]

I agree that we can try to guess the meaning by looking at what the OP has done; but the OP needs to realize that clearer communication is important.

Do you mean
$$\frac{\arctan(1+x)}{1-x}$$
or do you mean
$$\arctan \left( \frac{1+x}{1-x} \right)?$$
If you mean the latter you need parentheses, like this: arctan((1+x)/(1-x)) or arctan[(1-x)/(1+x)].

Its the second one , but in my original post that's exactly how I entered it.

 

[AlexandraMarie112]

I agree that we can try to guess the meaning by looking at what the OP has done; but the OP needs to realize that clearer communication is important.

Do you mean
$$\frac{\arctan(1+x)}{1-x}$$
or do you mean
$$\arctan \left( \frac{1+x}{1-x} \right)?$$
If you mean the latter you need parentheses, like this: arctan((1+x)/(1-x)) or arctan[(1-x)/(1+x)].

Okay I just noticed I actually put two brackets. I can see now why everyone is confused and guessing. Thanks for correction.