I. - For each of the following values of c, find lim_x->c+f(x), lim_x->c+f(x) and lim_x->c+f(x).

Refer to the graph on the test.

c	lim x->c+	lim x->c-	lim x->c
0	1	1	1
2	2	3	dne
4	3	2	dne

(Note that the limit exists iff the left-hand and right-hand limits are equal.)

II.

1.

lim x->0

(1)

=

1

2.

lim x->4

(x1/2)

=

2

3.

lim x->2

[(x2-4)/(x-2)]

=

lim x->2

[(x+2)]

=

4    (since x2-4=(x+2)(x-2))

4.

lim x->1

[(x1/2-1)/(x-1)]

=

lim x->1

1/(x1/2-1)

=

1/2    (since x-1=(x1/2+1)(x1/2-1))

5.	lim h->0	[((x+h)2-x2)/h]	=	lim h->0	[(x2+2xh+h2-x2)/h]
			=	lim h->0	[h(2x+h)/h]
			=	lim h->0	(2x+h)
			=	2x

6.	lim h->0	[(sin(x+h)-sin(x))/h]	=	lim h->0	[(sin(x)cos(h)+cos(x)sin(h)-sin(x))/h]
			=	lim h->0	(sin(x)[cos(h)-1]+cos(x)sin(h))/h)
			=	lim h->0	(sin(x)[(cos(h)-1)/h] + cos(x)[sin(h)/h])
			=	sin(x)*0 + cos(x)*1
			=	cos(x)

III. - Determine whether or not the function is continuous at the indicated point. If not, determine whether the discontinuity is a removable discontinuity, a jump discontinuity, or neither.

1. f(x)=(x³-1)(x-1), at x=1

Since f(x) is undefined at x=1 (division by zero), it can't be continuous there. However, since (x³-1)/(x-1) simplifies (after factoring the numerator) to x(x+1), the limit of f(x) as x->1 is 1(1+1)=2. Since this limit exists, f(x) has a removable discontinuity at x=1.

Note that a rational function is defined iff the denominator is not zero. This is the case here, so f(x) is continuous at x=0.

From the definition of h(x) for x < 2, we see that the left-hand limit of h(x) (as x->2^-) is 4, (See part II., #3) and the right-hand limit (as x->2⁺) is 2³=8. Since both limits exist, but are not equal, h(x) has a jump discontinuity at x=2.

(Note: since lim_x->2h(x) does not exist, the actual value of h(2) is irrelevant; no matter what it is, h(x) will have a jump discontinuity at x=2.)

IV. - In this section, consider the function f(x)=x³-x-4.

1. Use the Intermediate Value Theorem to show that f(x) has a root in the interval [1,2].

Since f(x) is continuous on the closed interval [1,2], the Intermediate Value Theorem tells us that, for any value y between f(1) and f(2), there will be some value of x in [1,2] such that y=f(x). (Note: to get full credit for this problem, you had to point out that f(x) is continuous on [1,2]; otherwise, the Intermediate Value Theorem would not apply.)

Since f(1)=-4<0 and f(2)=2>0, f(x)=0 for some x between 1 and 2 (i.e., 1 < x < 2).

The number of iterations necessary is n, where (0.5)ⁿ<0.1. That is:

• One bisection guarantees a result within 0.5 of the correct answer.
• Two bisections guarantee a result within 0.25 of the correct answer, etc.

It turns out that you need to bisect the interval 4 times to guarantee a result within 0.1 (actually within 0.0625) of the actual root of f(x). The results of your bisections should be, in order:

1. (1+2)/2=1.5;

f(1.5) = -2.125 < 0. (Thus, x is between 1.5 and 2.)
2. (1.5+2)/2 = 1.75;

f(1.75)=-0.390625 < 0. (Thus, x is between 1.75 and 2.)
3. (1.75+2)/2 = 1.875;

f(1.875)=0.7167... > 0. (Thus, x is between 1.75 and 1.875.)
4. (1.75+1.875)/2 = 1.8125. This is the 4^th bisection, so we may stop here.

Of course, more bisections will result in even closer approximations. The next four iterations give us: 1.78125, 1.796875, 1.7890625, 1.79296875.

Note: for those of you who work with computers often (comp sci majors, for example), a good exercise would be to write a computer program to quickly approximate the roots of a given function with the Bisection Method. Talk to me if you'd like to do this as an extra-credit assignment.

1. If f and g are both continuous at c, and g(c) 0, then (f/g) is also continuous at c.

First, rewrite what is given and what is to be shown in mathematical terms:

Given:	lim	f(x)	=	f(c),	and	lim	g(x)	=	g(c)
	x->c					x->c

Show:	lim	[f(x)/g(x)]	=	f(c)/g(c)
	x->c

Our starting point is what is given, and our finishing point will be whatever is to be shown. All that remains now is to write down the step(s) in between (i.e., the proof).

From what we are given, the following is simply substitution:

(	lim	f(x)) / (	lim	g(x))	=	f(c)/g(c)
	x->c		x->c

This is almost what we want to show (see above); however, the left-hand side isn't quite right. Fortunately, one of the Limit Theorems states that the limit of a quotient of two functions (which is what we want) is equal to the quotient of their limits (which we have). Therefore, by applying this Limit Theorem, the above equation becomes:

Show:	lim	[f(x)/g(x)]	=	f(c)/g(x)
	x->c

...which is exactly what we had to show. 

To show that something does not exist, a good strategy is to assume (i.e., "pretend") that is does exist, in order to get a contradiction. That is, if your assumption leads to a conclusion which isn't possible, then the assumption must not be true.

So for this problem, let us assume:

lim	f(x)	=	L,    for some real number L
x->0

(In other words, we assume that the limit exists, and thus has some real value, which we're calling L.) This gives us the following limit equation:

lim	xf(x)	=	(	lim	x)	(	lim	f(x))    (Note: this step is valid iff both limits exist.)
x->0				x->0			x->0
		=	0 * L
		=	0

BUT, in the original statement, we were given lim_x->0(xf(x))=1! Thus, we've shown that this limit is both 1 and 0, which of course is impossible!

We now have our contradiction, which was brought on by our assumption of the existence of lim_x->0(f(x)). Therefore, this limit does not exist.

Solution: see the proof in the textbook, section 2.5. (It should also be in your lecture notes.)

See the solution to IV.2. earlier on this page. The last value given there is within .004 of the actual root of f(x). Using a calculator or (preferably) a computer, you can approximate this root as closely as you like.

Note: it is possible to find the exact solution to this (or any other) cubic equation. This is because an analogue of the Quadratic Formula for polynomials of degree 3. This formula is known as Tartaglia's Solution; Tartaglia was a mathematician who lived in the 16th century. I don't know this formula; I only know that it exists and is rather complicated. If you're curious, you should be able to find it in the math library, or by doing a web search on "Tartaglia."

Answer: 1/(cos²(x)), or sec²(x).

This problem is actually similar to problem 6 in section II or this test, though it requires a few more steps. When we get to section 3.6 in the text, we will learn how to more easily differentiate tan(x) (as well as the other trigonometric functions).