About eigensteve Expertise Any questions about introductory or advanced calculus, real or complex analysis, linear algebra, or ordinary differential equations are fair game. I can answer questions about specific calculations, derivations, proofs, and physical applications. Your questions are as good for me as they are for you!
Experience I use linear algebra and differential equations almost every day for my research on modeling unsteady aerodynamics. In particular, my research experience includes numerical integration of trajectories, stability analysis of fluid flow fields, numerical computation of Lyapunov exponents, and more.
Organizations SIAM
AIP
APS
AIAA
IEEE
Publications AIAA Aerospace Sciences Meeting and Exhibit [2008, 2009].
IEEE Photovoltaic Specialist [2009].
For these publications, see http://carlsbad.princeton.edu/~steve/papers/
Education/Credentials I earned my B.S. in Mathematics from Caltech in 2006. I am currently a PhD. Candidate in Mechanical and Aerospace Engineering at Princeton. I expect to graduate in 2011.
Awards and Honors Athena-Feron Scholarship Award for Excellence in Mathematical Coursework [2007],
Princeton MAE Second Year Fellowship for Research Excellence [2007],
Gordon Wu Fellowship [2006-2010],
Caltech Summer Undergraduate Research Fellowship [2003-2005].
Question I am working on a proof that should be fairly simple but
keep getting stuck.
the problem is prove that
lim (1/x-1) does not exist
x-->1
Now what I did was assume the limit does exist and tried to
find a contradiction.
So for any number e>0 we can find a number d>0 such that
|f(x)-L|<e
whenever 0<|x-a|<d
now x=(d/2)+1 and x=(-d/2)+1 both satisfy the above
equation.
so I plug in (d/2)+1 and (-d/2)+1 into the equation for
|f(x)-L|<e
and I get |(1/((d/2)+1-1) - L|<e
and |(1/((-d/2)+1-1) - L|<e
which simplify to
|2/d - L|<e
and
|-2/d - L|<e --> which also equals |2/d + L|<e
picking an arbitrary value for e (like 1 or 1) and solving
for L does not seem to present me with a contradiction.
Can you help?
Answer Hi Richard,
The issue is in the condition "whenever 0<|x-a|<d". Given some specific e>0, we should be able to find a d so that /for all/ 0<|x-a|<d, the inequality |f(x)-L|<e holds.
However, in the example you chose with x=(d/2)+1, if d is chosen small enough, then clearly 2/d blows up.
