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    Suppose that (i) f(x) \rightarrow y_0 as x \rightarrow x_0, (ii) g(y) \rightarrow l as y \rightarrow y_0 and (iii) g(y_0) = l.

    Prove carefully that  g(f(x)) \rightarrow l as  x \rightarrow x_0

    Not sure how to do this.

    I would usually say simply that g(f(x)) tends to g(y0) as x tends to x0 but the carefully bit I presume means I can't just do this.
    I think I need to get to

    \forall \epsilon>0 \ \ \exists \delta >0 \ \ ,\ \ |g(f(x)) - g(y_0)|< \epsilon \ ,\  \forall x \ \ s.t \ \ \ 0<|x-x_0|<\delta

    but i'm not sure how to do it

    If g(y)->l as y->y0, then \forall \epsilon &gt; 0, \exists \delta_1 &gt; 0 s.t. |g(y)-g(y_0)|&lt;\epsilon for all y with 0<|y-y0|<delta_1

    But if f(x)->y0, then \forall \delta_1&gt;0, \exists \delta_2 &gt; 0 s.t. |f(x)-f(x_0)| &lt; \delta_1 for all x with 0 < |x-x0| < delta_2.

    Being careful...
    Note that we haven't used one of the conditions. The condition we haven't used *is* necessary. Try to work out why.
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Updated: January 22, 2010
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