Finding constant in a function

[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.

 

[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.

 

[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.

 

[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

 

[alex1261]

Yes! Thank you for you help!

 

[HallsofIvy]

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

 

[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

 

[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.

 

[hunt_mat]

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

 

[HallsofIvy]

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:
$$\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$$
Mat

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.

 

[hunt_mat]

Incorrect posting perhaps?