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Calculus/Assymptotes

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Expert: Alon Mandes - 6/9/2009


Question
How would I find the horizontal assymptotes algebraically for the function y=x/sqrt(x^2-4). I know from a graphing calculator that they are 1 and -1 but I cant seem to figure it out algebraically.  Thanks.

Answer
y=x/sqrt(x^2-4)
The vertical assymptotes are : x=2 & x=-2 .
They are calculated by asking when y goes to +INF & -INF.
Here's the solution :
sqrt(x^2-4)=0 -> x=2 & x=-2 .
The horizontal assymptotes are : y=1 & y=-1 .
They are calculated by asking what is the limit of y when x goes
to +INF & -INF .
Here's the solution :
Lim    x/sqrt(x^2-4) =  
x-> INF
Lim    1/sqrt(1- 4/x^2) =  1/sqrt(1) = 1 & -1 .
x-> INF

Alon.

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Alon Mandes

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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 .

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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".

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

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M.A in Mathematics & Bs.c in Electronics.

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