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Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lessons 1 and 2: Did you read my tips on how to study and learn Math 1500? If not, here is a link to those important suggestions: Did you read my tips I sent earlier on making sign diagrams and finding range? Here are the links: Did you see my tips for Assignment 1? These are tips for the assignments in the Distance/Online Math 1500 course, but I strongly recommend that you do this assignment as homework even if you are taking the classroom lecture section of the course. These assignments are very good (and challenging) practice. The
first assignment is a great way to build and review key skills that will be helpful for this course. Here is a link to the actual assignment, in case you don't have it: Study Lesson 2 (Limits), Lesson 3 (Continuity), and Lesson 4 (The Definition of Derivative) from my Intro Calculus book to prepare for this assignment.
Part (a)
You can only compute the logarithm of a positive number, so ln0 does not exist, nor does ln(negative). Memorize that, as x approaches 0+, ln(x) heads to negative infinity. However, as x approaches 0-, ln(x) does not
exist since you cannot log a negative number. Consequently, this limit cannot be approached from both sides. This is kind of similar to the situation you deal with in square root of 0 limits that I discuss in Lesson 2.
Parts (b) and (c)
I show you how to do limits similar to these as part of my Lesson 3. Break them up into one-sided limits (0- and 0+)
then use the appropriate piece of the piecewise function to solve each one-sided limit. As always, if both one-sided limits agree, the limit exists. If they disagree, the limit does not exist.
Parts (d) and (e)
These are pretty simple examples of square root limits. Only if you have square root of 0 is there anything to investigate. That makes part (e) pretty
trivial.
Parts (f) and (g)
As I review in Lesson 5, a negative exponent is really telling you that there is a denominator. So they are giving you 1/x^4 and 1/x^5, respectively. That creates k/0 limits which I discuss in Lesson 2.
Parts (h) and (i)
Classic 0/0 limits. Be
careful, sometimes after, you factor and cancel, a 0/0 limit turns into a k/0 limit. See my Lesson 2 Practise Problems 47 and 48.
Part (j)
Classic conjugate 0/0 limit, like my Lesson 2, question 2.
Part (k)
Similar to my Lesson 2, question 16(c). Be sure you have
studied all about trig limits and done all of question 16, starting on page 60 of my book. Remember, Lesson 2 is included in my free samples above if you haven't got my book. (How could you not get my book, if that is the case?)
This is a Squeeze Theorem question. Study all of Lesson 2, question 17 starting on page 63 of my book. Start by saying -1 <= sin(ln|x|) <= 1.
This is more of the same stuff you dealt with above in question 1 of this assignment. Now, they are just giving you a one-sided limit to do.
This is a runthrough of all the concepts involved in doing Limits at Infinity as discussed in Lesson 2, questions 10-15 in my book. Be sure you have also looked at my Practise Problems 57-74. Especially question
74.
Some additional facts to learn:
- ln(x) heads to infinity as x approaches infinity.
- As mentioned above, you can't do ln(negative) or ln(0), but ln(x) approaches negative infinity as x approaches 0+.
- e to the power of x will approach infinity as x approaches infinity. However, it will approach 0 as x approaches negative infinity (because that is actually creating a negative exponent, and so ultimately creating
1 divided by infinity). I discuss this in Lesson 9 of my book.
- Again, just like Assignment 1, they have introduced inverse trig even though that is not part of the course syllabus. Do note that the inverse tangent function (which is what they mean by tan with the -1 superscript given in part (i)) gives three important values that you may want to memorize (although, why should you, since this isn't supposed to show up
until Math 1700).
- inverse tan(0) = 0; or, tan^-1 (0) = 0
- inverse tan(x) = pi/2 as x approaches infinity because tan(pi/2) takes off to infinity.
- inverse tan(x) = negative pi/2 as x approaches negative infinity.
These are similar to my Lesson 3, questions 1-3.
Note, for the first one, where f(x) = absolute value of x+1, you should redefine that as a piecewise function using the concepts about the absolute value function I first discuss in
Lesson 2, question 4. A quick sign diagram should demonstrate that the zero is at x= -1, and that x+1 is negative for x<-1 and positive for x>-1. Therefore, you know that absolute value of x+1 is -(x+1) for x<-1 and it is simply x+1 for x>-1. Of course, when x=-1, you get absolute value of 0, which is 0. You could then define a three-piece piecewise function for f(x) using similar notation to my examples in Lesson 3, or what they have done for
h(x) in this question. Tell them what f(x) is for x<-1, x=-1, and x>-1 using the piecewise notation, then proceed to examine the continuity.
Again, similar to my Lesson 3, questions 1-3, especially question 3.
This is just a long-winded way of asking you to compute f'(x) using the definition of derivative as I teach in Lesson 4. In Lesson 5, I discuss the fact that velocity is the derivative of
distance.
Be careful that you use proper notation. Instead of f'(x), you should be finding s'(t). So, say that v(t), the velocity at any time t, is s'(t) where s'(t) is the limit as h approaches 0 for s(t+h) - s(t) all divided by h. Essentially, you are doing exactly what I do in my Lesson 4, question 1(b) except, you are using s instead of f, and t instead of x.
More like my Lesson 4. Do note that the power of -1/2 is saying square root in the denominator. I go into more detail about all of this in Lesson 5. You should rewrite this as f(x) = 1/ sqrt(1+2x), and then proceed. Quite
similar to my Lesson 4, question 2(b).
Note that these are actually the same functions for which you already investigated continuity back in question 5 of the assignment. Consequently, since you are expected to know the theorem Differentiability implies Continuity (this is the very first theorem
proof you have to memorize, as summarized on page 3 of my book, Required Proof #1), anywhere that you proved the function is discontinuous in question 5, immediately proves that is also not differentiable at that x value. This will be helpful for h(x).
Be clear, Differentiability implies Continuity tells us:
- If f(x) is discontinuous at x=a, then it is certainly not differentiable at x=a
either. Which is to say, f'(a) does not exist.
- However, if f(x) is continuous at x=a, that does not necessarily mean f(x) is differentiable at x=a. It is still possible that f'(a) does not exist. You must investigate that further.
I think you should be able to declare that polynomials are differentiable. Therefore, each piece of the given piecewise functions (and make sure you have redefined the absolute value function f(x)
as a piecewise function as I discussed in my tips for question 5 above), can be declared to be differentiable for their OPEN intervals (i.e. up to but not including the endpoint of each interval) since each piece is merely a polynomial.
However, at the x-value where the pieces meet (that's x=-1 in the case of f(x), and x=0 in the case of g(x) and h(x)), you must investigate if the derivative exists.
To compute
f'(-1), you must use the definition of derivative, but do the limit as h approaches 0- and the limit as h approaches 0+ as two separate limits. I suggest, rather than do f(x+h) - f(x) on top, sub -1 in place of x immediately, so do f(-1+h) - f(-1) all divided by h. But, be very careful to use the correct piece of f:
- As h approaches 0-, that would be equivalent to x approaching -1-, so use the correct piece of f(x) for when x<-1.
- As h approaches 0+,
that would be equivalent to x approaching -1+, so use the correct piece of f(x) for when x>-1.
After you have computed those two one-sided limits, compare your answers. If you got different answers (and you better! one should be -1 and the other should be 1), you have proven the limit as h approaches 0 does not exist, therefore f'(-1) does not exist. This proves f(x) is not differentiable at x=-1.
You have to do something similar for
g(x) and, perhaps, h(x).
Hint: g(x) should be proven to be differentiable at x=0, i.e. g'(0) does exist, since the limits for h approaches 0- and 0+ will have the same answers, even though you should have used very different pieces for each one.
Of course, h(x) is not differentiable at x=0 since you should have proven back in question 5 that h(x) is not continuous at
x=0.
Similar to my Lesson 3, questions 4-5, especially question 4.
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