1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limits of functions, inequalities (analysis)

  1. Jan 22, 2010 #1
    1. The problem statement, all variables and given/known data

    1) f(x) [tex]\rightarrow[/tex] yo as x [tex]\rightarrow[/tex] x0
    2) g(y)[tex]\rightarrow[/tex] l as y [tex]\rightarrow[/tex] y0
    3) g(y0) = l

    Prove that g(f(x)) [tex]\rightarrow[/tex] l as x [tex]\rightarrow[/tex] x0

    2. Relevant equations

    Definition of function tending to limit - E is a subset of R, f:E->R, f tends to l as x tends to p if for all e>0 there exists a d>0 such that |f(x) - l | < e for all x in E such that 0 < |x-p|< d.

    Theorem: f: E [tex]\rightarrow[/tex] R where E [tex]\subseteq[/tex] R, p is a limit point of E and l [tex]\in[/tex] R. The following are equivalent:
    a) f(x) [tex]\rightarrow[/tex] l as x [tex]\rightarrow[/tex] p
    b) for every sequence {pn} in E such that pn [tex]\neq[/tex] p and lim[tex]_{n\rightarrow\infty}[/tex] pn = p we have that f(pn) [tex]\rightarrow[/tex] l as n[tex]\rightarrow[/tex] [tex]\infty[/tex]

    3. The attempt at a solution

    I thought I could use the above theorem with (1) to say that every sequence {xn} such that xn[tex]\neq[/tex] x0 and lim[tex]_{n\rightarrow\infty}[/tex] xn = x0, we have that f(xn) [tex]\rightarrow[/tex] y0 as n [tex]\rightarrow[/tex] [tex]\infty[/tex]

    Similarly from (2) every sequence {yn} such that yn[tex]\neq[/tex] y0 and lim[tex]_{n\rightarrow\infty}[/tex] yn = y0, we have that g(yn) [tex]\rightarrow[/tex] l as n [tex]\rightarrow[/tex] [tex]\infty[/tex]

    So, can I let yn = f(x) ( or maybe f(xn)?
    Would this help show that g(f(x)) [tex]\rightarrow[/tex] l? I haven't used (3) or really shown this happens as x [tex]\rightarrow[/tex] x0.
  2. jcsd
  3. Jan 22, 2010 #2
    You don't need sequences to do this; you can use the definition of limit directly. Find a condition on [tex]y[/tex] that ensures that [tex]g(y)[/tex] is close to [tex]l[/tex], then find a condition on [tex]x[/tex] that ensures that [tex]f(x)[/tex] satisfies that condition on [tex]y[/tex].
  4. Jan 23, 2010 #3
    From (3) I know for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]1 > 0 such that |g(y)-l|<[tex]\epsilon[/tex] for all y such that 0<|y-y0|<[tex]\delta[/tex]1

    Similarly from (1)
    for all [tex]\epsilon[/tex]>0 [tex]\exists[/tex] [tex]\delta[/tex]2 > 0 such that |f(x)-y0|<[tex]\epsilon[/tex] for all x such that 0<|x-x0|<[tex]\delta[/tex]2

    Can I then do something along these lines:

    =|g(f(x)) - g(y0) -l|
    =|g(f(x)) - 2l|

    Is less than [tex]\epsilon[/tex] for all [tex]\delta[/tex]=min([tex]\delta[/tex]1, [tex]\delta[/tex]2) such that 0<|x-x0|<[tex]\delta[/tex]?
  5. Jan 23, 2010 #4
    They are not correct because one can't ensure that g(f(x)-y0) is close to [tex]l[/tex] and of course g(y) is not a linear function to assign such a property to it that g(f(x)-y0) = g(f(x))-g(y0). Try to apply both the f(x) and g(f(x)) to the triangular inequality!

  6. Jan 23, 2010 #5
    How is this?

    Let [tex]\epsilon[/tex]> 0
    [tex]\exists[/tex][tex]\delta[/tex]1>0 such that |g(y)-l| = |g(y) - g(y0| < [tex]\epsilon[/tex]
    [tex]\exists[/tex] ][tex]\delta[/tex]2>0 such that |f(x) - y0| < [tex]\delta[/tex]2
    So for all x such that |x-x0|< min([tex]\delta[/tex]1,[tex]\delta[/tex]2) we have
    |(gof)(x) - (gof)(x0)| = |g(f(x))-g(f(x0))| = |g(f(x)) - l| < [tex]\epsilon[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook