What value of a will make this continuous at x=0?

Question

The function f is given by the formula f(x)=(2x)/x^2+6x when x<0

and by f(x)=cos(x)+a when x>(or equal to)0

What value must be chosen for a  in order to make this function continuous at x=0?

Answer 1

Continuity has three conditions that you must meet. If you use these three conditions, all continuity problems become the same.

For your problem, we want to test the continuity at x = 0 for f(x)

1) Does f(0) exist?

Yes, since at x = 0, f(0) = cos(0) + a

So, f(0) = 1 + a

2) Does lim f(x) exist.

x→0

Based on the way that f(x) is defined, you can see that f(x) has values of x defined on the left of x = 0 (I.e. x < 0) and to the right of zero. (I.e. x ≥ 0) *Note: the equal sign could've been on either one of these, but this is just how your problem was defined.

Because of the way f(x) is defined in the problem, we'll check the left and right side limits. And if the limit does exist, these two one - sided limits must be equal to each other.

lim f(x) = lim (2x)/(x² + 6x) <------- I chose this function, because for x < 0, f(x) is defined as 2x/(x² + 6x)

x→0^- x→0^-

= lim (2x)/[x(x + 6)] <----------Factor out an x in the denominator.

x→0^-

= lim 2/(x + 6) <----------- Take the limit as x approaches zero now.

x→0^-

= 2/6 = 1/3

Now for the right hand side.

lim f(x) = lim (cos(x) + a) <------- f(x) is defined as cosx + a for x ≥ 0 (which is on the right side of zero.

x→0⁺ x→0⁺

= cos(0) + a

= 1 + a

Finally, if the lim f(x) exists, then

x→0

lim f(x) = lim f(x)

x→0^- x→0⁺

1/3 = 1 + a

a = -2/3

*We can stop right here, as we just found a. However, just for the sake of completeness, condition 3 states

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3) lim f(x) = f(0)

x→0

So, the limit value should be 1/3 (because the left and right side should both be the same if the limit exists) the left sided limit is 1/3 and the right sided limit is 1 + a = 1 + (-2/3) = 1/3

f(0) = 1 + a = 1 + (-2/3) = 1/3

Thus, we have 1/3 = 1/3 so condition 3 is satisfied.

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Hope this helps

Mr. K

Answer 2

For x ≠ 0

f(x) = 2 / (x + 6)

As x → 0, f(x) → 1/3

f(x) = cos x - 2/3