uni calculus exam question (1 Viewer)

[underthesun]

Sorry, I guess this question might not fit in the forum, but after thinking about the kind of viewers to this..

Just a q : in an exam class test, when it asks "evaluate", do we have to prove everything? for example:

evaluate lim _(x->3) 2x + 4 and such.

esp these keywords: determine, find, state

argh calculus exam tomorrow.. and I thought i'd be free of cramming like in the HSC days..

example exams without example answers

 

[Xayma]

Evaluate effectively means "Just put me in your calculator and give me the answer" ie you dont have to prove anything.

For example if you evaulate an odd function with ends a and -a the answer will be 0.

 

[McLake]

Originally posted by underthesun
Just a q : in an exam class test, when it asks "evaluate", do we have to prove everything? for example:

...

esp these keywords: determine, find, state

As Xayma said, evaluate means "find an answer". The same goes for the other keywords ...

 

[Estel]

Heh in an exam of mine in a 2 mark evaluate question I lost a mark because I didn't show any working... even though I felt it was just a basic calculator question.

 

[underthesun]

Originally posted by McLake
As Xayma said, evaluate means "find an answer". The same goes for the other keywords ...

Whoa, so what was all the calculus lecturers been talking about lately then? (i.e limits, the epsilon symbol, and the delta symbol)

 

[Ragerunner]

Didn't want to make a new thread so I'll ask a question in here.

Let f(x) = sqroot(x) / (sqroot(x) + 5)

For each є > 0, find an M(є ) such that f(x) is within є of lim {x -> infinity} f(x) whenever x > M(є )

I'm totally stumped

 

[Affinity]

sun: thought you would post the test questions up

rage:
hmm... let M(e) = 25/e^2

if x > M(e)

then:
x> 25/e^2
sqrt(x) > 5/e
sqrt(x) + 5 > 5/e
e > 5/[sqrt(x) + 5]
e > | -5 / [ sqrt(x) + 5] |
e > | sqrt(x)/[sqrt(x) + 5] -1 |


ofcourse the hard part is to find M(e) in the first place

 

[redslert]

Originally posted by Affinity
hmm... let M(e) = 25/e^2

if x > M(e)

then:
Underthesun: I thought I was going to see the exam questions for tomorrow.

Ragerunner:
x> 25/e^2
sqrt(x) > 5/e
sqrt(x) + 5 > 5/e
e > 5/[sqrt(x) + 5]
e > | -5 / [ sqrt(x) + 5] |
e > | sqrt(x)/[sqrt(x) + 5] -1 |


ofcourse the hard part is to find M(e) in the first place

did u just make up the bloody numbers as you go?!

huh???

btw i failed my calculus test today la di da

 

[Affinity]

oh.. I inserted the message to underthe sun in the wrong place..

beh.. I need to study , mine is coming up 16:00 tomorrow

 

[redslert]

hmmmm affinity why the heck did u work backwards??

from the defintion
let є >0
then |f(x) - L|<є whenever |x-a| < delta
bah bah

so you want to find є

but you have to work out the limit first L at infinity

lim (x->infinity) f(x) = 1

.'. |f(x) - 1|<є

|sqrt(x)/[sqrt(x) + 5] - 1| < є
and then you work it out from there

but you seem to have gone backwards?! huh?

 

[Xayma]

Originally posted by underthesun
Whoa, so what was all the calculus lecturers been talking about lately then? (i.e limits, the epsilon symbol, and the delta symbol)

Is the delta symbol the small delta (ie looks like a curvy d or the capitol delta), the small delta represents an extremley small amount, ie just above something else, for example when testing a point of inflexion at x³ you could test 0-delta x and 0+delta x remembering that any integer is bigger then delta x and delta x² is smaller then delta x. Otherwise it means change

 

[Affinity]

I did it forwards...

you are supposed to find d not e, remember it's epsilon - delta, not delta-epsilon.

the way you have it is backwards and if teachers are picky.. they will require you to start with the M and prove |f(x) -L| < e if
0<|x-a|<d

 

[...]

Originally posted by Affinity
let M(e) = 25/e^2


ofcourse the hard part is to find M(e) in the first place

:whoa:

but how did u pick e??
like..wtf

 

[redslert]

Originally posted by Affinity
I did it forwards...

you are supposed to find d not e, remember it's epsilon - delta, not delta-epsilon.

the way you have it is backwards and if teachers are picky.. they will require you to start with the M and prove |f(x) -L| < e if
0<|x-a|<d

hmmmmmm not according to my lecture notes and also my tutor!!
the way that i stated above is the way that i was shown...hmmm

otherwise how would anyone be able to just come up with
25/e^2

i mean, even if you were the luckiest person on earth....you can't just come up with that.....and it actually works??

so yeh.....how did u know that 25/e^2 will work?????????

 

[...]

hmm..thats the MQ method too..well, i think it is..lol

like u do a | f(x) - 1 | thing and multiply the conjugate??

 

[redslert]

multiply conjugates eh hmmm

that depends on the question

what your trying to do is just make x by itself...if you had to multi by conju...to get x
then you do it

but for the above question it isn't required

 

[Ragerunner]

Originally posted by redslert
btw i failed my calculus test today la di da

I just did it today....

*jumps off cliff*

MAN SO BLOODY HARD WTF....

some questions from my test

1) Find the values of a and b where

f(x) =

ax + b [x <= 1]
tan(pi/4) [1 < x < 2]

is differentiable at 1.


2) find the limit {x -> infinity } (2x + 3x + e^-2x) / (5x - cosx + sinx)

something similar to that equation, i can't remember it exactly, how would i find limits to infinity of equations such as those which have a lot of values?

3) find lim {x -> 0 } x^1/3 * sin(1/x)


thanks.

 

[Xayma]

I think 3) is 0 as -1<= sin (1/x) <=1
and x^1/3 approaches 0 as x approaches 0

 

[Affinity]

2.) compare dominant terms of numerator and denominator.

we have in general as x>infinity

factorial > exponentials > polynomials/positive powers > linear > logarithmic > constant

with polynomials/powers

if a>b>0
then x^a > x^b

 

[maniacguy]

What you're *supposed* to do is work out that the limit is 1, because you can see it's 1 (sqrt(x)/(sqrt(x)+5) clearly goes to 1).

Then you work out the conditions necessary on delta to have |f(x)-L|< eps whenever |x-a|<delta.

Affinity, your method is technically also right, but because M(eps) is usually much harder to see intuitively than the limit L, usually it's done the other way, so as not to look completely arbitrary (I bet you worked out your M(eps) using the method suggested by the others )

In a way you're pulling the answer (the limit) out of thin air and then checking to see that it's right by seeing if you can find a delta. If you can't find any delta = M(eps) to make the thing work then either your working is wrong or the limit you found is wrong.