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Assume that $\underset{x\rightarrow b}{\lim}f(x)$ exists.

Show that $\forall a,b\in\mathbb{R}$, if $a\neq 0$ then we have: $$ \underset{x\rightarrow0}{\lim}f(ax+b)=\underset{x\rightarrow b}{\lim}f(x)$$

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    $\begingroup$ Put $t=ax+b$ then as $x\to 0$ we have $t\to b$ and then you are done. $\endgroup$ – Paramanand Singh Dec 5 '15 at 10:25
  • $\begingroup$ Use the definition of composite functions $\endgroup$ – Hedwig Dec 5 '15 at 10:38
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Suppose $\lim_{x\to b}f(x) = L$ exists. Then given $\varepsilon > 0$, there is $\delta > 0$ such that $$ \lvert x-b\rvert < \delta \implies \lvert f(x)-L\rvert < \varepsilon. $$ Let $\delta_1 = \dfrac {\delta} {\lvert a\rvert}$. Then if $\lvert x\rvert = \lvert x-0\rvert < \delta_1$, we have $$\lvert ax+ b - b\rvert = \lvert ax\rvert = \lvert a\rvert \lvert x\rvert < \lvert a\rvert \delta_1 = \delta,$$ thus $$ \text{for all $x$, if }\lvert x\rvert < \delta_1 \text{ then } \lvert f(ax+b)-L\rvert < \varepsilon. $$ So $\lim_{x\to 0} f(ax+b) = L$.

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Composites of continuous functions are continuous. This means that if $ f(g(x)) $ is defined on an interval containing $c$, and $ \lim_{x\to c} g(x)=L$, then: $$ \lim_{x\to c} f(g(x))=f(L)=f\big(\lim_{x\to c} g(x)\big) $$ From this your kan solve your question by: $$ \lim_{x\to 0} f(ax+b) = f(\lim_{x\to 0}(ax +b)) = f(b)=\lim_{x\to b} f(x)$$ Since $ \lim_{x\to b} f(x)$ exists.

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    $\begingroup$ $f$ is not assumed to be continuous, or even continuous at $b$. The limit exists, but it's not stated $x\to b$ means inclusive or exclusive of $b$. It could be exclusive of $b$!, and then $f(b) \ne $ the limit. $\endgroup$ – BrianO Dec 5 '15 at 10:44

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