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Prove if the following is correct or not: If $$\lim _{x\to x_0}f(x) = L \text{ and } \lim_{x\to x_1}g(x) = x_0,$$ then $$\lim_{x\to x_1} f(g(x))= L.$$ So, I guess this can be solved either by proving it or find a example that contradicts the above, so this statement is wrong.

Michael Hardy
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Belf
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    Have you looked up the relevant definitions? If not: look up the definition of a limit (start with the most basic definition for $\mathbb{R}$). Apply the definition twice. Tell us more precisely where you get any problems. – Andrey Tyukin Oct 26 '15 at 18:51
  • As in epsilon-delta? – Belf Oct 26 '15 at 18:55
  • Yes, the epsilon-delta definition would be a reasonable starting point. I don't think that you need the more general versions for metric/topological spaces. – Andrey Tyukin Oct 26 '15 at 18:56
  • Well, from the first limit we have for every epsilon>0, there is delta>0 so that | f(x) - L | > 0. And from the second limit we have | g(x) - xo | > 0. I'm stuck here :P – Belf Oct 26 '15 at 19:02

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For all $\varepsilon>0$ there exists $\eta>0$ such that whenever $0<|x-x_0|<\eta$ then $|f(x)-L|<\varepsilon$.

Given $\eta>0$, there exists $\delta>0$ such that whenever $0<|x-x_1|<\delta$ then $|g(x) - x_0|<\eta$.

So $\underbrace{0<|x-x_1| <\delta \Longrightarrow |g(x)-x_0|<\eta \Longrightarrow |f(g(x)) - L|<\varepsilon}_\text{This falls just a little bit short.}$.

The difficulty here is that we don't have $0<|g(x) -x_0|<\eta$. That suggests that if $g(x) = x_0$, there might be a problem. Consider the case where $g(x) = x_0$ regardless of the value of $x$. Then $g$ is a constant function. Suppose then that $f(x) = \dfrac{x^2 - x_0^2}{x-x_0}$. This simplifies to $x+x_0$ when $x\ne x_0$, but is undefined when $x=x_0$. It limit as $x\to x_0$ is $L=2x_0$. If $g$ is constantly equal to $x_0$, then $\lim_{x\to x_1} g(x) = x_0$ but $f(g(x))$ would be undefined, and thus cannot approach $L$.

However, if $f$ is continuous at $x_0$, then the conclusion is correct and the argument above is almost valid. The only thing one would need to change to make it valid is to say that if $f$ is continuous at $x_0$ and $\lim_{x\to x_0} f(x) = L$ (which, in this case of continuity, implies $f(x_0)=L$), is that we would need to say the following: $$ \text{For all }\varepsilon>0 \text{ there exists } \eta>0 \text{ such that whenever } |x-x_0|<\eta \text{ then } |f(x)-L|<\varepsilon. $$

Michael Hardy
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  • This is a very detailed answer, however, I don't think that one should care about the cases where $x_0$ is not in the domain of $f$... OP has not specified what space he is working with, so I'd propose to assume that $x_0$ is in the domain of $f$. – Andrey Tyukin Oct 26 '15 at 21:04
  • @AndreyTyukin : If $x_0$ is in the domain of $f$, that doesn't mean $\lim_{x\to x_0} f(x) = f(x_0)$. It is given that $\lim_{x\to x_0} f(x)$ exists; it is not given that $f$ is continuous at $x_0$. ${}\qquad{}$ – Michael Hardy Oct 26 '15 at 21:50
  • I read it as follows: $\lim_{x\to x_0}f(x) = L$ means in particular that $f(x_0) = L = \lim_{x\to x_0}f(x)$, because the constant sequence $x_k = x_0$ is a possible choice for a sequence that converges to $x_0$. If we wanted to enforce $x\neq x_0$, we would have to write something like $\lim_{x\to x_0+}$ or $\lim_{x\to x_0-}$ or $\lim_{x\to x_0, x\neq x_0}$ instead. So I assume that $f(x_0)=L$. – Andrey Tyukin Oct 26 '15 at 21:56
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    The statement $\lim_{x\to x_0} f(x) = L$ is by definition based on values of $f(x)$ for $x\ne x_0$. My answer explains how to get the conclusion in the case where $f$ is continuous at $x_0$, i.e. in the case where $\lim_{x\to x_0} f(x) = f(x_0)$. ${}\qquad{}$ – Michael Hardy Oct 26 '15 at 22:00
  • Hm, apparently there are different conventions? You assumed the "deleted limit"-definition, I assumed the "non-deleted limit" definition. The problem is not your or mine choice of definition, the problem is that OP didn't say anything about the domains and codomains of the functions, and also left unspecified what exactly (s)he means by the limit notation. – Andrey Tyukin Oct 26 '15 at 22:12
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    Limits are by definition "deleted". If $f:D\to E$ then the value of $\lim_{x\to x_0} f(x)$ is determined entirely by the behavior of $f$ on $D \setminus {x_0}$. ${}\qquad{}$ – Michael Hardy Oct 26 '15 at 22:15
  • According to Wikipedia (https://en.wikipedia.org/wiki/Limit_of_a_function#Deleted_versus_non-deleted_limits), "deleted limits are the most popular". I cannot tell how representative those five listed books are. This convention has almost no practical relevance anyway, because $x_0$ is either $\infty$, or outside of the domain of $f$, or $f$ is continuous or cadlag or something else, so one can always infer the meaning from the context. I apologize for the choice of the wrong convention, and swear to conform to The Deleted Limit Convention in those few cases where it actually does matter. ok? – Andrey Tyukin Oct 26 '15 at 22:52