Question

2. Let f: R R be a continuous function. Suppose that f is differentiable on R\{0} and that there exists an L e R such that lim,of,(z) = L. Prove that f is differentiable at 1-0 with f,(0) = L. (Hint: Use the definition of derivative and then use mean value theorem)

Answer 1

Since f is continuous on $\mathbb{R}$ and differentiable on $\mathbb{R}$ \{0} so f is continuous on $[0 , \frac{1}{n}]$ and differentiable on $(0 , \frac{1}{n})$ for all $n \in \mathbb{N}$

Hence by mean value theorem for all x , y $\in (0 , \frac{1}{n} )$ there exist $\xi \in (0 , \frac{1}{n} )$ such that ,

f(x) - f(y) = f'( $\xi$ ) (x - y)

Take x = h and y =0 then ,

f(h) - f(0) = f'( $\xi$ ) h

Or , $\frac{f(h) - f(0)}{h} = f'(\xi)$

Now as $\bigcap_{n=1}^{\propto} [0 , \frac{1}{n})$ ={ 0}

So , $\lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h} = \lim_{h\rightarrow 0 } f'( \xi)$

Or , $\lim_{h\rightarrow 0} \frac{f(h) - f(0)}{h} = L$ , using given condition .

Since the limit exist so f is differentiable at 0 and since the limit equals to L .

Hence f'(0) = L

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