Question

Determine the limits, if it exists.
1) limx->0 abs(3x-1)-abs(3x+1) / x
2) lim x->0 square root 9+x^2 - square root 9-x^2 / x^2

Answer 1

What have you done so far?

Answer 2

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@ganeshie8 @agent0smith

Answer 3

@IsTim works too :)

Answer 4

Let's see what has been attempted first, but have you considered trying the conjugate for the second?

Answer 5

for the 1) i open up the absloute sign and get rid of the 3x which i got abs -2 / x and then sub in the zero 2/0 ... so it doesn't exist. is that correct?

Answer 6

Are you allowed to use L'hopital?

Answer 7

i dont know what's l'hospital

Answer 8

https://www.khanacademy.org/math/differential-calculus/derivative_applications/lhopital_rule/v/introduction-to-l-hopital-s-rule

Answer 9

http://mathworld.wolfram.com/LHospitalsRule.html

Answer 10

If you don't know it then I assume it's calc I.. so I guess, yeaa, you'd be right on your answer.. let's wait for IsTim tho.

Answer 11

what will i do the second question with square roots? i don't know how to approach it

Answer 12

For 1)

lim [x, 0]: (abs(3x-1) - abs(3x+1) ) /x

Well, abs(3x-1) can be split into two seperate conditions to get rid of the absolute value signs:

3x - 1 if 3x-1 > 0
and
-(3x -1) if 3x -1 < 0

solving for x, in 3x -1, we get x = 1/3.

So we can rewrite our conditions

3x - 1 if x > 1/3
and
-3x + 1 if x < 1/3

Our limit, x approaches 0, meaning we will always be less than 1/3.

So we can rewrite abs (3x-1) to simply -3x +1.

Using the same logic and method for abs(3x+1), it yields we can rewrite as 3x+1.

So our limit that we need to eval is now:

lim[x, 0]: ((-3x +1) - (3x+1))/ x
simplify

lim[x,0]: -6x/x

lim[x,0]: -6

Answer 13

oh i see. thanks, itstim

Answer 14

For the second one, you would multiply top and bottom by the conjugate which is sqrt(9-x^2) + sqrt(9+x^2)

This simplifies your limit to:
(9+x^2 - (9-x^2))/ ((x^2) * (sqrt(9-x^2) + sqrt(9+x^2))

Simplifying this further you get
(2x^2)/ ((x^2) * (sqrt(9-x^2) + sqrt(9+x^2))

Simplify..
2 / ((sqrt(9-x^2) + sqrt(9+x^2))


Now just plug in our limit value, 0

and you get 2/6