evinda
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Hello! :D
I am looking at this exercise:
Let f: (-a,a)\rightarrow \mathbb{R} continuous at 0.We suppose that for a t \epsilon (0,1) this condition is satisfied:
\lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}=b
Show that f'(0) exists and find its value.
I thought that I could write it like that \lim_{x\rightarrow 0}\frac{f(x)-f(0)+f(0)-f(tx)}{x}=b \Rightarrow \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}+\lim_{x\rightarrow 0}\frac{f(0)-f(tx)}{x}=b\Rightarrow (1-\frac{1}{t})\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}=b
So,f'(0) exists and is equal to \frac{bt}{t-1} .Are my thoughts right or am I wrong?
I am looking at this exercise:
Let f: (-a,a)\rightarrow \mathbb{R} continuous at 0.We suppose that for a t \epsilon (0,1) this condition is satisfied:
\lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}=b
Show that f'(0) exists and find its value.
I thought that I could write it like that \lim_{x\rightarrow 0}\frac{f(x)-f(0)+f(0)-f(tx)}{x}=b \Rightarrow \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}+\lim_{x\rightarrow 0}\frac{f(0)-f(tx)}{x}=b\Rightarrow (1-\frac{1}{t})\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}=b
So,f'(0) exists and is equal to \frac{bt}{t-1} .Are my thoughts right or am I wrong?