Thread: Piecewise and Trig Limits

This one has got me totally stumped:


I've been told to find the values of a and b so f(x) is continous everywhere. I'm not exactly sure of the method used to find the two integers in order to make the piecewise statement true, and therefore continous.

Also, I have these 2 to do as well:



I'll probably kick myself whenever I figure the bottom two out. I recall the process being fairly simple, but when you haven't slept in about a day and a half, your brain starts to become a little fuzzy.

Some help on these would be great... I've been working on this worksheet since 12:00.

1)
So that there is continuity, the middle branch of the function, f(x) = ax +b, must connect with the first and third branches of the function.
So get the point where the 1st branch ends, and the point where the 3rd branch begins, then solve for the middle branch using those two points.

The end of the 1st branch is at x = -1, or, since the 1st branch is not defined at x = -1, the 1st branch tends to finish at x = -1.
So,
f(x), as x --> -1, = 2(-1)^2 -3(-1) +1 = 6.
That's point (-1,6).

The start of the 3rd branch is at x = 2,
So, f(2) = 8(2) +5 = 21
That's point (2,21)

The line that will connect those points has a slope of
m = (21 -6) / (2 -(-1)) = 15/3 = 5
Its equation is, by point-slope form,
y -(6) = 5(x -(-1))
y = 5x +5 +6
y = 5x +11
And that should be the f(x) = ax +b
Hence, a = 5, and b = 11 -----------------answer.

------------------------------------

Lim(x -->0) of tan(4x) / 6x
= 0/0

Use the L'Hopital's rule
= Lim(x --> 0) of sec^2(4x) *4 / 6
= (1)*4 / 6
= 2/3 ----------answer.

-------------------------------
Lim(x --> 0) of [(3x -sin(x))^2] / x^2
= 0/0

Use the L'Hopital's rule.
= Lim(x --> 0) of [2(3x -sin(x))(3 -cos(x)] / 2x
= 0/0

Use the Rule again,
= Lim(x --> 0) of 2[(3x -sin(x))(sin(x)) +(3 -cos(x))(3 -cos(x))] / 2
= 2[0 +(3 -1)(3 -1)] / 2
= 4 -------------------------answer.