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About 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".

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

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

 
   

You are here:  Experts > Teens > Homework/Study Tips > Calculus > limit question

Calculus - limit question


Expert: Alon Mandes - 7/11/2008

Question
Hi,
I have a question similar to
lim x approaches infinity of e^-x times e^x.

It seems to me that the limit here would be 1, but the graph I have indicates the answer is 0.
How do I get started actually working this problem?

Answer
Ok, our function is e^(-xe^x). Let's get started first by
intuition: e^(-xe^x) can be written as [e^(-x)]^(e^x) which means(1/e^x)e^x. As we know, e^x grows rapidly, thus 1/e^x declines rapidly as well .(because it becomes fraction smaller than 1), & further more, "this fraction" is powered by e^x which itself grows rapidly. Now we conclude that our function will decline very very rapidly to zero. (Fraction)^x -->0 , (fraction)^(e^x)--->0 faster.

Now let's consider our claim analytically : Let's set y=e^(-xe^x), we
perform Ln for both sides, we get : Ln(y)=-xe^x. We can agree that
-xe^x goes to -(infinity) when x goes to infinity. Our question will be now : Which value of y can make the Ln function reaches the
-infinity ??? & the answer is 0. It's pretty obvious from the graph of the Ln function. In other words when x reaches infinity, y reaches zero. & y is exactly our function.

Summery : Lim (x-->inf) e^(-xe^x) = 0.

Alon.  

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