Calc. Limit questions..

nein12 · #1 $Old$ June 17th, 2009, 05:09 PM

Need some help with these annoying questions. If any one of you guys have time to take a look and give me a shout, it'll be great. I am taking online course, so it takes a while for me to get my tutor.

find lim x-> 3 for f(x)
f(x):
1/(x-3)^2 if x doesn't equal 3
2 if x=3

I don't think there is a limit because x-> 3- has undefined value... but I am not sure.

lim x-> 0
sqaure root (x-x^2)

I could say it's 0, but I think it doesn't have one because function itself doesn't exist when x is less than 0, and so there's no lim x-> 0-... hence no overall function.

lim x -> 1+
sqaure root (1-x)
Same reasoning as earlier question

f(x) x-5 / sinx.. give interval notation
I know x does not equal n(pi) where n is any integer... but how do I express that in interval notation?

Thanks in advance,

nein12

Jhevon · #2 $Old$ June 17th, 2009, 05:54 PM

Quote:

Originally Posted by nein12 $View Post$

Need some help with these annoying questions. If any one of you guys have time to take a look and give me a shout, it'll be great. I am taking online course, so it takes a while for me to get my tutor.

find lim x-> 3 for f(x)
f(x):
1/(x-3)^2 if x doesn't equal 3
2 if x=3

I don't think there is a limit because x-> 3- has undefined value... but I am not sure.

$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^-}f(x) = \infty$

so...

Quote:

lim x-> 0
sqaure root (x-x^2)

I could say it's 0, but I think it doesn't have one because function itself doesn't exist when x is less than 0, and so there's no lim x-> 0-... hence no overall function.

lim x -> 1+
sqaure root (1-x)
Same reasoning as earlier question

by convention, both these limits are zero

Quote:

f(x) x-5 / sinx.. give interval notation
I know x does not equal n(pi) where n is any integer... but how do I express that in interval notation?

Thanks in advance,

nein12

hmm, depends on what you are allowed to do. some answers may look like:

$\bigcup (n \pi, (n + 1) \pi)$ for all integers $n$

or

$(-\infty, \infty) \backslash n \pi$ for all integers $n$

or

$(0, \pi) \cup (-\pi, 0) \cup (\pi, 2\pi) \cup (-2 \pi, \pi) \cup \cdots (n \pi, (n + 1) \pi) \cup (-(n + 1) \pi, - n \pi) \cdots$ for all $n$

nein12 · #3 $Old$ June 17th, 2009, 06:06 PM

Jhevon,

Thanks a lot for your time and effort. I have a bit of trouble grasping some of your answers...

Quote:

Originally Posted by Jhevon $View Post$

lim x-> 0
sqaure root (x-x^2)

I could say it's 0, but I think it doesn't have one because function itself doesn't exist when x is less than 0, and so there's no lim x-> 0-... hence no overall function.

lim x -> 1+
sqaure root (1-x)
Same reasoning as earlier question

so...

by convention, both these limits are zero

I could substitute the x variables with zero to get that, but I used the graphing calculator to double check this answer. From what I see, the graph domain for square root (x-x^2) is 0 < or equal x < or equal 1. As you see, 0 is at the very left end of the function.

In order for limit as x->0 to exist, x->0+ and x->0- must be equal to each other. x->0+ exists because function from 1 to 0 is continuous. x->0- however cannot be continuous as any values less than 0 does not exist.

Hence, limit for x->0+ exists (0, of course) and limit for x->0- is undefined. Hence, the overall limit does not exist.

Similar argument for the square root (1-x) as x -> 1+

Am I just being paranoid, or is there a flaw in my arguement? Again, I would appreciate if you could get back to me.

nein12

Jhevon · #4 $Old$ June 17th, 2009, 06:11 PM

Quote:

Originally Posted by nein12 $View Post$

Jhevon,

Thanks a lot for your time and effort. I have a bit of trouble grasping some of your answers...

I could substitute the x variables with zero to get that, but I used the graphing calculator to double check this answer. From what I see, the graph domain for square root (x-x^2) is 0 < or equal x < or equal 1. As you see, 0 is at the very left end of the function.

In order for limit as x->0 to exist, x->0+ and x->0- must be equal to each other. x->0+ exists because function from 1 to 0 is continuous. x->0- however cannot be continuous as any values less than 0 does not exist.

Hence, limit for x->0+ exists (0, of course) and limit for x->0- is undefined. Hence, the overall limit does not exist.

Similar argument for the square root (1-x) as x -> 1+

Am I just being paranoid, or is there a flaw in my arguement? Again, I would appreciate if you could get back to me.

nein12

your reasoning is sound. but like i said, it is convention. if you type either of those limits into some CAS, for instance, they will give you zero as the answer.

nein12 · #5 $Old$ June 17th, 2009, 06:23 PM

Quote:

Originally Posted by ;330905

your reasoning is sound. but like i said, it is convention. if you type either of those limits into some CAS, for instance, they will give you zero as the answer.

Hmmm... I guess I have to find out why that's conventional. Nevertheless, thanks a bunch for your help.

VonNemo19 · #6 $Old$ June 17th, 2009, 06:47 PM

For a limit to exist, it has to be able to be approached from both sides; unless of course we are only concerned about taking a one-sided limit.

This is implied by the definition.

$\mid{f(x)-L}\mid<\epsilon\text{ whenever }\mid{x-c}\mid<\delta$

The absolute value signs provide the implication.

nein12 · #7 $Old$ June 21st, 2009, 10:06 PM

Quote:

Originally Posted by Jhevon $View Post$

your reasoning is sound. but like i said, it is convention. if you type either of those limits into some CAS, for instance, they will give you zero as the answer.

I did as you said, and I lost the marks. (Good thing it was a practice test) My tutor pointed out exactly the same reasoning as I did. If I was a tutor, how could you justify that?

Jhevon · #8 $Old$ June 22nd, 2009, 10:50 AM

Quote:

Originally Posted by nein12 $View Post$

I did as you said, and I lost the marks. (Good thing it was a practice test) My tutor pointed out exactly the same reasoning as I did.

as i said, to me your reasoning was sound, but it seemed strange that they were giving you all these limits that don't exist, so i decided to double check with a CAS. both matlab and maple gave answers for the limits, so i assumed it was convention. similar to how you can say $\lim_{x \to 0} \sqrt x = 0$ , even though the function is not defined to the left of zero. it seems that if the function is continuous at the point in question, the limit is taken to be that value. i suppose you were expected to be more strict and use the $\epsilon-\delta$ definition of the limit. sorry you lost the marks.

nein12 · #9 $Old$ June 22nd, 2009, 11:30 AM

Quote:

Originally Posted by Jhevon $View Post$

as i said, to me your reasoning was sound, but it seemed strange that they were giving you all these limits that don't exist, so i decided to double check with a CAS. both matlab and maple gave answers for the limits, so i assumed it was convention. similar to how you can say $\lim_{x \to 0} \sqrt x = 0$ , even though the function is not defined to the left of zero. it seems that if the function is continuous at the point in question, the limit is taken to be that value. i suppose you were expected to be more strict and use the $\epsilon-\delta$ definition of the limit. sorry you lost the marks.

It was a practice test, so it's ok. But I guess this is a lesson for me to question everything... even if it's from math experts. I kinda lied to myself and kinda went with "conventional" excuse, even if I didn't understand fully.

But thanks anyway. Just gotta be a bit more careful from now on... it doesn't help that they ask me really weird questions. Typically, they ask you simple equation questions.. some worded questions and applications... but this institution ask awfully a lot about theoretical questions and rules.