# Finding constant in a function

1. Aug 20, 2010

### alex1261

Determine all values of the constant a such that the following function is continuous for all real numbers.

f(x) = ax/tan x, if x≥ 0
a^2 -2, if x<0

I've tried to plug in zero but the tan x throws me off because plugging in zero will give me a 0/0. Other than that I am stuck on what to do.

2. Aug 20, 2010

### CompuChip

So the limit from the left as you approach zero is clearly a2 - 2.
For the limit from the right, you get 0/0, which is an indeterminate form. Therefore, you can use for example L'Hopitals rule to try and find the limit.

3. Aug 20, 2010

### alex1261

Hi can you explain the L'Hopitals rule? I'm sorry I'm just starting Calculus and my teacher assigned this problem and we have yet to learn it. Thanks.

4. Aug 20, 2010

### hunt_mat

To avoid L'Hopital's rule, note that:
$$\tan x=x+\frac{x^{3}}{3}+\cdots$$
so:
$$\frac{ax}{\tan x}=\frac{a}{1+x^{2}/3+\cdots}$$
Then the limit as x tends to zero is equal to a, so from here, can you answer the rest of the question?

Mat

5. Aug 20, 2010

### alex1261

Yes! Thank you for you help!

6. Aug 20, 2010

### HallsofIvy

I would be very surprized if a class had covered power series expansions of functions but not L'Hopital's rule!

7. Aug 20, 2010

### hunt_mat

If you can't do L'Hopital's rule or power series, how can you evaluate the limit? I suppose that if you were shown that:
$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$
by some other method, and you know about the algrebra of limits, then you could evaluate by brute force I suppose.

Mat

8. Aug 20, 2010

### Char. Limit

You're not going to find a constant that makes this continuous for all real numbers. The cotangent function is discontinuous at an infinite number of points, and multiplying it by x won't solve that problem.

9. Aug 20, 2010

### hunt_mat

I think he means continuous at the point x=0...

10. Aug 21, 2010

### HallsofIvy

That was my thought.
$$\frac{ax}{tan(x)}= a\frac{x}{sin(x)}cos(x)$$
Since both cos(x) and x/sin(x) go to 1 as x goes to 0, this has limit "a". $a^2- 2$ is a constant so to have this function continuous at x= 0, we must have $a^2- 2= a$ which is equivalent to $a^2- a- 2= (a- 2)(a+ 1)= 0$. That has the two roots a= 2 and a= -1.

However, I note that the original post specifically said "continuous for all real numbers" and, as Char. Limit said, no matter what a is, the function is not continuous for x any multiple of $2\pi$ except 0.

11. Aug 22, 2010

### hunt_mat

Incorrect posting perhaps?