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Expert: Alon Mandes - 10/15/2008


Question
I understand limits perfectly I m a genious at math, but how to find limits on absolute values stumps me. Here's the question find the limit as x approaches 3 when y=|x-3|/(x-3). please help

Answer
Whenever there is absolute, it means there is a shift of sign. So
the trick here is to know that :
|x-3|=x-3 if (x-3)>0 & |x-3|=3-x if (x-3)<0
We will calculate both limits :
1. Approaching 3 from the right (" x->3{R} ")
2. Approaching 3 from the left (" x->3{L} ")
Ok, let's get to work :

Lim       |x-3|/(x-3)    =
x->3{R}
Lim        (x-3)/(x-3)    =  
x->3{R}
Lim         1   =  1   (" because x-3{R} > 0 ")
x->3{R}

Lim       |x-3|/(x-3)    =
x->3{L}
Lim       (3-x)/(x-3)    =
x->3{R}
Lim         -1   =  -1  (" because x-3{L} < 0 ")
x->3{R}

We got that both limits are not equal !! & that means the limit
at 3 do not exist.

Alon.

Calculus

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Alon Mandes

Expertise

Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

Experience

1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

Education/Credentials
M.A in Mathematics & Bs.c in Electronics.

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