Calculus problem?

Question

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

http://img525.imageshack.us/img525/526/mathxq4.png

i dont have the faintist idea on how to do this

Answer 1

Basically, you figure what value of "a" will make the two equations equal at 0.

I graphed the two equations, and plugged in a random small number for a. I got lucky and found that it's 3.

Answer 2

I would think the problem is not so much with a, but with the split at 0. 1/tan x becomes infinite at zero, so the inequality should be revised to exclude 0 from the allowable values. Moreover, 1/tan x becomes infinite at pi, 2pi, etc, so those points have to be excluded. The only way I see to have a continuous function is to have abs (Tan x) in the function.

Answer 3

for the function to be continuous, it must have the same limit from both sides, that is lim as x_>0- = lim as x-> 0+...

we can factor out the a(since a is a constant...)

a^2-2 = a(lim (x/tan x) [as x approaches 0 from the right...])

a^2-2 = a (lim (x(cos x)/(sin x))
a^2-2 = a (lim (cos x)/((sin x)/x))

we know that lim as xapproaches 0 from the right of cos x = 1, and lim as x approaches 0 from the right of (sin x)/x = 1 as well... so if we get the limit, we will get...

a^2-2 = a

a^2-a-2 = 0

(a-2)(a+1)

a = -2, a = 1

Answer 4

find the limit of each using the [(x+h) -f(a)]/h. from the left and from the right. and once you find the limit then set each equal to each other and solve for a