Suppose the function $f :\mathbb{R}\to\mathbb{R}$ has limit $L$ at $0$, and let $a > 0$. If $g :\mathbb{R}\to\mathbb{R}$ is defined by $g(x) := f(ax)$ for $x\in \mathbb{R}$. Show that $\displaystyle{\lim_{x\to 0} g(x) = L}$.
Asked
Active
Viewed 134 times
2 Answers
1
Let $\varepsilon > 0$. We wish to show that there is some $\delta$ such that $|f(ax) - L| < \varepsilon$ whenever $|x| < \delta$.
Well, we know that there is some $\hat \delta$ such that $|f(x) - L| < \varepsilon$ whenever $|x| < \hat \delta$, so if we take $\delta = \frac{\hat \delta}{a}$ then we have $|ax| < \hat \delta$ if $|x| < \delta$, which implies that $|f(ax) - L| < \varepsilon$ and we are done.
MT_
- 19,603
- 9
- 40
- 81
0
Let $\varepsilon > 0.$ Since $f(x) \to L$ as $x \to 0,$ there is a $\delta' > 0$ such that $|f(x) - L| < \varepsilon/2$ for all $|x| < \delta'$ and there is a $\delta'' > 0$ such that $|f(ax) - f(x)| < \varepsilon/2$ for all $|x| < \delta''.$ So there is a $\delta > 0$, say $\delta := \min \{ \delta', \delta'' \},$ such that $$|f(ax) - L| = |f(ax) - f(x) + f(x) - L| \leq |f(ax) - f(x)| + |f(x) - L| < \varepsilon$$ for all $|x| < \delta.$
Yes
- 20,719