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let function $f$ at $x=a\in \mathbb{R}$ be differentiable and $n ,m , k \in \mathbb{R}$

then prove that : $$\lim\limits_{h \to 0} \frac{f(a+mh)-f(a+nh)}{kh}=\frac{m-n}{k}f'(a) $$

My Try :

since function $f$ at $x=a\in \mathbb{R}$ is differentiable : So we have :

$$f'(a)=\lim\limits_{x \to a} \frac{f(x)-f(a)}{x-a}=L \in \mathbb{R} $$

Now : $x-a= h$ :

$$f'(a)=\lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}=L \in \mathbb{R} $$

Now what ?

Almot1960
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1 Answers1

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Hint:

$$ \begin{align} \lim_{h \to 0} \frac{f(a+mh)-f(a+nh)}{kh} & = \lim_{h \to 0} \frac{f(a+mh)-f(a)- \big(f(a+nh)-f(a)\big)}{kh} \\[5px] & = \frac{m}{k} \, \lim_{h \to 0} \frac{f(a+mh)-f(a)}{mh} - \frac{n}{k} \, \lim_{h \to 0} \frac{f(a+nh)-f(a)}{nh} \end{align} $$

dxiv
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