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Let f be a function. Let g be a continuous function.

Consider the following limits:

$\underset{x\rightarrow x_{0}}{lim}f(g(x))$

$\underset{y\rightarrow g(x_{0})}{lim}f(y)$

I want to know if the existence of one of these limits imply the existence of the other.

Moreover, if both of these limits exist will they be necessarily equal ?

My gut-feeling is that I have to reason with the explicit definition of limits, so:

Suppose that $\underset{x\rightarrow x_{0}}{lim}f(g(x))=L_{1}$ then:

$\forall\epsilon_{1}>0,\exists\delta_{1}>0$ such that $|x-x_{0}|<\delta_{1}\rightarrow|f(g(x)-L_{1}|<\epsilon_{1}$

Somehow, from this I have to conclude that $\underset{y\rightarrow g(x_{0})}{lim}f(y)=L_{2}$ that is:

$\forall\epsilon_{2}>0,\exists\delta_{2}>0$ such that $|y-g(x_{0})|<\delta_{2}\rightarrow|f(y)-L_{2}|<\epsilon_{2}$

or the other way around ie existence of $L_{2}$ implies existence of $L_{1}$.

Any suggestions?

popololvic
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  • You might find my answer here useful: https://math.stackexchange.com/questions/2631325/substitution-in-finding-limits/2647916#2647916 – zhw. Mar 02 '18 at 17:21

1 Answers1

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If $g$ is the constant function $0$, and $f(x)=1$ for $|x|>0$, $f(0)=-1$, then $\lim_{y\rightarrow 0}f(y)=1$ and $\lim_{x\rightarrow x_{0}}f(g(x))=\lim_{x\rightarrow x_{0}}f(0)=-1$.

For the existence, let $f(x)=\chi_{\bf{Q}}(x)$, the function taking value $1$ on rational numbers and zero otherwise, and let $g$ be the constant $0$, then $\lim_{x\rightarrow 0}f(g(x))=\lim_{x\rightarrow 0}f(0)=1$ and $\lim_{y\rightarrow 0}f(y)$ does not exist.

Another example is like, let $g(x)=x\sin(1/x)$ for $x\ne 0$, $g(0)=0$, and $f(y)=1$ for $y\ne 0$, $f(0)=-1$, then $\lim_{y\rightarrow 0}f(y)=1$ exists, and the $\lim_{x\rightarrow 0}f(g(x))$ does not exist because for $x_{n}=\dfrac{1}{2n\pi}$, $f(g(x_{n}))=f(0)=-1$ but for $z_{n}=\dfrac{1}{(2n+(1/2))\pi}$, $f(g(z_{n}))=f\left(\dfrac{1}{(2n+(1/2)\pi)}\right)\rightarrow 1$.

user284331
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