# Prove $\underset{x\rightarrow0}{\lim}f(ax+b)=\underset{x\rightarrow b}{\lim}f(x)$

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: $$\underset{x\rightarrow0}{\lim}f(ax+b)=\underset{x\rightarrow b}{\lim}f(x)$$

• Put $t=ax+b$ then as $x\to 0$ we have $t\to b$ and then you are done. – Paramanand Singh Dec 5 '15 at 10:25
• Use the definition of composite functions – Hedwig Dec 5 '15 at 10:38

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 $\lim_{x\to 0} f(ax+b) = L$.
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.
• $f$ 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. – BrianO Dec 5 '15 at 10:44