Is this right?

Question

Canyou plz tell me if i got these problems correct, bc my friend here sais i didnt, but i dont think he is right...help lpz?

#1
lim ( square_root(6-h) - square_root(6) ) / h
h-->0

#2
lim ********** absolute_value( x + pi ) / ( x + pi )
x--> -pi

#3
lim ( (1/(5-x)) - (1/5) ) / x
x-->0


Ok, for the problem #1 and #2 my answer is that the limit does not exist and for #3 the limit is 1/25 . Can you tell me if these answers are correct OR show me how to do them if they are wrong? thx so much!

Ok im geting different answers from different people...could u plz answer only if u r sure u are rigt? thx

ok DISREGARD the ******
Here is another way to write problem #3

lim [ (1/5-x) - (1/5) ] / x
x-->0

Answer 1

I noticed that you were very confused about your answers. I will try to help you. Im sorry for the other guys, it is not my intention to publish your mistakes, but to help the Sir-excalibur. And to help you too, why not? Please dont get mad at me.

ABOUT THE REMARKS THAT YOU PUT BELOW THE QUESTION:

It is not the same

lim [ (1/5-x) - (1/5) ] / x
x-->0

than

#3
lim ( (1/(5-x)) - (1/5) ) / x
x-->0


ABOUT OTHER ANSWERS:
youwishyou: Since you werent right, his answer is wrong

captain Aha B

Hes almost right about the derivative, but not about the conclusion. This is similar to the derivative of - sqrt (x) evaluated in x = 6, so you get -1/(2sqrt6)

f`(a) = lim [f(a+h)- f(a)]/h, when h tends to 0

Or, considering -h ->0, you could write this this way:

f`(a) = lim [f(a-h)- f(a)]/(-h), when - h tends to 0

So, the expression that you have would be -f`(a)

But you cant use this to conclude than the result is 0 because it is a constant.

Yo yo is not right. Lilian is not right, either. h is not 0, h tends to 0. Actually 6-h must be greater or equals than 0, but, since we are calculating the limit in a neighbourhood from 0, then 6-h is a positive number

The same for x different from pi, etc.

Im sorry, Liliam, if you check out the definition of
limit you will notice that you are working in a neighbourhood from 0 BUT NOT IN 0

The persons who suggested l`Hopital were right

gth is not right in her results.

Doug-Donahue and rfamilymembers answer #1 are positive and they should be negative. The other answers from the first are wrong, rfamilymembers answer are right (but the #1, because of the - sign)


Ana

Answer 2

#1 Use this:
(a+b)(a-b) = a^2 = b^2

lim ( square_root(6-h) - square_root(6) ) / h
h-->0

Multiply and divide by
square_root(6-h) + square_root(6)

So you will get this:

lim (6-h-6)/[(sqrt(6-h) + sqrt (6))*h]
h-->0


lim -h/[(sqrt(6) + sqrt (6)) * h]
h-->0

lim -1/2sqrt 6 = -1/2sqrt(6)
h->0

#2 abs A = A sign A

So abs A/ A = sign A

Sign A = 1 if A is positive, -1 is A is negative, and 0 is A is 0.

I need to know if x->- pi+ or x -> pi-

If you want a limit for x-> -pi, the answer is: there is not a limit


#3 lim ( (1/(5-x)) - (1/5) ) / x
x-->0

lim [5 - (5-x)]/[(5-x)5]x =
x-> 0

lim x/5.5.x = 1/25
x->o

Answer 3

first is the derivative of sqrt(6). this is a constant, so the derivative is zero.

Actually - check the parentheses - I'm not sure about how you stated the problem. if the h is outside the limit, then the answer is zero - if inside, I think it is - infinity.

2) The ***** confuses me - what does that mean?

This is the same as lim( abs(y)/y) as y goes to zero

from the left, it is -1, from the right, it is 1 therefore, the limits don't agree and this llimt doesn't exist.

3) Check your parentheses on this and 1) - not sure you didn't mean to include the x and h in the ().

In this case, the lim of the right term (-1/5/x) is -infinity, so yeah, doesn't exist

Answer 4

If you haven't learned L'Hospital's Rule yet, then all three DNE.

#3 becomes
((1/(5-0))-(1/5))/0

Top is 1/5 - 1/5
Bottom is 0

becomes 0/0

But with LH you got 3 right by me...

#3 dy/dx (5-x)^-1 is -(5-x)^-2 * -1
dy/dx 1/5 is 0
dy/dx x is 1 so it's
((5-x)^-2)/1 the limit as x-->0 of which is
(1/25)/1 = 1/25

bah i will check the others
1 LHed is (-.5(6-h)^(-.5))/1
limit is (1/(sqrt(6))*(-1/2) which is what? (-1)/(2sqrt(6))? Bad at the simplifying stuff
so that one is wrong unless I totally f'ed up which is possible

Answer 5

alright, now its pretty late here where i am, and i havent done calc for alomst 3 months now, but i did get a perfect score on my calc ap exam in may...and because im so tired, all i could do was the first one:

because the limit as h approaches 0 is 0/0, the easiest way is to use l'hopital's rule, which is where you take the first derivative of the top and put it over the first derivative of the bottom. f' of the top i got to be -1/(2squareroot(6-h)) and f' of the top i got to be 1. so that leaves a final answer of -1/2root6. sorry if it's wrong, but thats the best i can do right now.

Answer 6

1.expand(6-h)^1/2 using the binomial theoremand proceed
6^1/2(1-h/6)^1/2
[6^1/2(1+1/2*h/6+...................)
taking limits
lim h tending to zero 6^1/2(h/12)/h=6^1/2/12
2.let x->-pi+h the limit will be=!-pi+h+pi!/(-pi+h+pi)=+1
let x->-pi-h the limit will be =!-pi-h+pi!/(-pi-h+pi)=-1
since the left and right limits are not the same the limit does not exist
taking LCM the expression=(5-5+x)/x(5)(5-x)
cancelling the x es out (thisis possible as x is not zero,it only tends to zero)
lim x->0 1/5(5-x)=1/25

Answer 7

apply L'Hopital's rule: that states that if the product of two functions tends to zero or infinity as it approaches a certain value, than take the derivative of that fuction and take the limit of that.


1. lim {(sqrt (6-h) - (sqrt(6)/h)}
------i think the limit is undefined

2. I think the limit is 1.

3. The limit is -1

Answer 8

(1)
The limit exists and is equal to 1/(2√6)

(2)
The limit exists and is equal to 1

(3)
The limit exists and is equal to 0

Now *you* go back and crack your textbook and figure out why.


Doug

Answer 9

DNE for all three. Anything divided by zero does not exist, even for #3.

Answer 10

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