if $\lim_{x \to 0} \frac{f(x)}{x} = l$ and $b\neq0$, then $\lim_{x \to o} \frac{f(bx)}{x} = bl$

Question

How to show USING an $\epsilon-\delta$ argument this:

If $\lim_{x\to 0} \frac{f(x)}{x} = l$ and $b\neq0$, then $\lim_{x \to 0} \frac{f(bx)}{x} = bl$

Thanks in advance.

Answer 1

$\lim_{x\to 0} \frac {f(x)}{x} = l$ means that for all $\epsilon >0$ there is a $\delta_{\epsilon} > 0$ so that for all $x; |x - 0| < \delta_{\epsilon}$ then $|\frac {f(x)}x - l| < \epsilon$.

So for $\frac {\epsilon}b > 0$ let $\delta_{\epsilon/b}$ be such that $| x- 0| < \delta_{\epsilon/b}$ implies $|\frac {f(x)}x - l| < \frac {\epsilon}{b}$.

Now let $\Delta = b*\delta_{\epsilon/b}$. Let $g(x) = \frac {f(x)}{x}$.

Then if $|x - 0| < \Delta$ then

$|x - 0| < b*\delta_{\epsilon/b}$. And

$|\frac xb - 0| < \delta_{\epsilon/b}$. And

$|\frac {f(xb)}{xb} - l| < \frac {\epsilon}b$. And

$|\frac {f(xb)}x-bl| < \epsilon$. And

$|g(x) - bl| < \epsilon$.

So $\lim_{x\to 0} g(x) = \lim_{x\to 0} \frac {f(bx)}b = bl$.

So let $\delta_{\epsilon} = \Delta$ we have proven that for all $x; |x-0|<\delta_{epsilon}$ then $|\frac {f(bx)}{x} - bl| < \epsilon$.

SO by definition $\lim_{x\to 0} \frac {f(xb)}x = bl$.

Answer 2

Given that $\lim_{x\to 0} \frac{f(x)}{x} = l$ we have

$$\left| \frac{f(x)}{x} - l\right|<\frac{\epsilon}{b} \quad \forall0<|x|<\delta$$

then

$$\left| \frac{f(bx)}{bx} - l\right|<\frac{\epsilon}{b}\quad \forall0<|bx|<\delta$$

and

$$\left| \frac{f(bx)}{x} - bl\right|<\epsilon \quad \forall0<|x|<\frac{\delta}{|b|}=\delta_1$$

Answer 3

1)Let $\epsilon/|b| >0$ be given.

There is a $\delta > 0$ such that

$|y|<\delta$ implies

$|f(y)/y -l| < \epsilon/|b|$.

2) Set $y:=xb$ then :

$|y|=|xb| \lt \delta$ implies

$|f(y)/y-l| = |f(xb)/xb -l| =$

$ |1/b| |f(xb)/x -bl| \lt \epsilon/|b|$, or

$|f(xb)/x-bl| < \epsilon.$