True or false on certain limit "theory" questions

s3a · #1 $Old$ November 20th, 2009, 06:49 PM

The following are the questions I am having trouble understanding. (The answers in the boxes are the correct answers)

"(1 pt) Enter a T or an F in each answer space below to indicate whether the corresponding statement is true or false. A statement is true only if it is true for all possibilities. You must get all of the answers correct to receive credit.

1. If

is continuous at

, then

is differentiable at

2.

3. If

and

, then

does not exist
4. If

and

, then

does not exist
5.

"

I have the answers but I was hoping someone could at least briefly explain why each one is true or false.

Any help would be greatly appreciated!
Thanks in advance!

ANDS! · #2 $Old$ November 20th, 2009, 06:56 PM

Put some spaces between those questions - they are bleeding into each other and I have crappy eyes.

Are you having trouble answering all of the problems?

tonio · #3 $Old$ November 21st, 2009, 04:04 AM

Quote:

Originally Posted by s3a $View Post$

The following are the questions I am having trouble understanding. (The answers in the boxes are the correct answers)

"(1 pt) Enter a T or an F in each answer space below to indicate whether the corresponding statement is true or false. A statement is true only if it is true for all possibilities. You must get all of the answers correct to receive credit.

1. If

is continuous at

, then

is differentiable at

2.

3. If

and

, then

does not exist
4. If

and

, then

does not exist
5.

"

I have the answers but I was hoping someone could at least briefly explain why each one is true or false.

Any help would be greatly appreciated!
Thanks in advance!

Next time do "preview" your answer to avoid cumbersome-looking posts as the above...

1) False: counterexample $f(x)=|x|$ at zero

2) True since both limits exist and are finite and the denominator's limit isn't non-zero

3) False: counterexample $f(x)=g(x)=x-4$ .

4) True, IFFFFF by "doesn't exist" you people mean the limit doesn't exists FINITELY. For example, $\lim_{x\to 4}\frac{x+1}{(x-4)^2}=\infty$ exists in the generalized sense. It all depends on your definitions.

5) Obviously false since then you'd get $\frac{0}{0}$ ...The actual limit is 7/5 .

Tonio

s3a · #4 $Old$ November 21st, 2009, 11:11 AM

Sorry for not previewing. Also thanks for the answer, but could you just confirm the following (in my own words) for me please?:

#3. the limit = 1 therefore, it does exist? (because something over same thing = 1)
#4. the limit = 5/0 therefore, it does not exist?

tonio · #5 $Old$ November 21st, 2009, 11:18 AM

Quote:

Originally Posted by s3a $View Post$

Sorry for not previewing. Also thanks for the answer, but could you just confirm the following (in my own words) for me please?:

#3. the limit = 1 therefore, it does exist? (because something over same thing = 1)

You don't have "something over the same" here, but two things that converge to the same AND DIFFERENT from zero, thus the limit exists and it's possible to evaluate it the way it's done in the question.

#4. the limit = 5/0 therefore, it does not exist?

You can't write that since $\frac{anything}{0}$ isn't defined in mathematics. What you can say is that while the numerator approaches a finite number as $x\xrightarrow [x\to 1] {}1$ , the denominator approaches $\infty$ and then you cannot try to evaluate the limit the way they're showing in the question.

Tonio