Show that f'(0) exists and find its value

[evinda]

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?

 

[alyafey22]

If the first derivative exists at $0$ then by L'Hosiptal rule we have

\(\displaystyle \lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}= f'(0)-tf'(0) = b \)

\(\displaystyle f'(0)= \frac{b}{1-t}\)

 

[evinda]

If the first derivative exists at $0$ then by L'Hosiptal rule we have

\(\displaystyle \lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}= f'(0)-tf'(0) = b \)

\(\displaystyle f'(0)= \frac{b}{1-t}\)

And that is the only thing I have to say?So...because of the fact that \(\displaystyle \frac{b}{1-t}\epsilon \mathbb{R}\) ,\(\displaystyle f'(0) \) exists,and its value is \(\displaystyle \frac{b}{1-t} \) ?

 

[chisigma]

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?

The mean value theorem extablishes that, given a function f(*) continuous in [a,b] and differentiable in (a,b), it exists a number c in (a,b) such that... $\displaystyle f^{\ '} (c) = \frac{f(b) - f(a)} {b - a}\ (1)$

Setting in (1) b=x and a= t x, if ...

$\displaystyle \lim_{ x \rightarrow 0} \frac{f(x) - f(t\ x)}{x} = \alpha\ (2)$

... is also...

$\displaystyle \lim_{ x \rightarrow 0} \frac{f(x) - f(t\ x)}{x\ (1 - t)} = \frac{\alpha}{1 - t} = f^{\ '}(0)\ (3)$

View attachment 1784

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[evinda]

The mean value theorem extablishes that, given a function f(*) continuous in [a,b] and differentiable in (a,b), it exists a number c in (a,b) such that... $\displaystyle f^{\ '} (c) = \frac{f(b) - f(a)} {b - a}\ (1)$

Setting in (1) b=x and a= t x, if ...

$\displaystyle \lim_{ x \rightarrow 0} \frac{f(x) - f(t\ x)}{x} = \alpha\ (2)$

... is also...

$\displaystyle \lim_{ x \rightarrow 0} \frac{f(x) - f(t\ x)}{x\ (1 - t)} = \frac{\alpha}{1 - t} = f^{\ '}(0)\ (3)$

And which interval [a,b] do we take??

 

[chisigma]

And which interval [a,b] do we take??

If You chose a= t x and b = x, then b-a = x (1 - t) and t x < c < x. If x tends to 0, then a, b and c tend to 0 and starting from...

$\displaystyle \lim_{x \rightarrow 0} \frac{f(x) - f(t\ x)}{x} = \alpha\ (1)$

... You arrive at...

$\displaystyle f^{\ '} (0) = \frac{\alpha}{1 - t}\ (2)$

View attachment 1785

Merry Christmas from Serbia

$\chi$ $\sigma$

 

[evinda]

If You chose a= t x and b = x, then b-a = x (1 - t) and t x < c < x. If x tends to 0, then a, b and c tend to 0 and starting from...

$\displaystyle \lim_{x \rightarrow 0} \frac{f(x) - f(t\ x)}{x} = \alpha\ (1)$

... You arrive at...

$\displaystyle f^{\ '} (0) = \frac{\alpha}{1 - t}\ (2)$

And $$[a,b]=[tx,x]$$ should be $$\subseteq (-a,a)$$ .Right?

 

[evinda]

And $$[a,b]=[tx,x]$$ should be $$\subseteq (-a,a)$$ .Right?

So,can I start like that:f is continuous in [a,b]$$\subseteq (-a,a)$$ and differentiable in (a,b),so it exists a number c in (a,b) such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$.
We select $$a=tx$$ and $$b=x$$ . ?

 

[Opalg]

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.

ZaidAlyafey's comment shows that if $f'(0)$ exists then it is equal to $\dfrac b{1-t}.$ The snag with this, and with all the other comments, is that you are not told that $f$ is differentiable. You are not even told that it is continuous, except at the single point $0$. So you cannot use l'Hôpital's rule, or the mean value theorem, or any other result that requires differentiability, and you need to prove that $f$ is differentiable at $0$.

You are told that $$\lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}=b$$. So given $\varepsilon>0$ there exists $\delta>0$ such that \(\displaystyle \left|\frac{f(x)-f(tx)}{x} - b\right| < \varepsilon\) whenever $|x|<\delta.$ Therefore $$|f(x) - f(tx) - bx| < \varepsilon|x|.$$ Next, if $|x|<\delta$ then $|t^nx|<\delta$ for all $n\geqslant1$ (because $t<1$). So in the above inequality we can replace $x$ by $tx$, $t^2x,\ldots, t^{n-1}x,$ to get $$|f(tx) - f(t^2x) - btx| < \varepsilon t|x|,$$ $$|f(t^2x) - f(t^3x) - bt^2x| < \varepsilon t^2|x|,$$ $$\ldots$$ $$|f(t^{n-1}x) - f(t^nx) - bt^{n-1}x| < \varepsilon t^{n-1}|x|.$$ Now add those inequalities and use the triangle inequality, to get $$\left|\sum_{k=0}^{n-1}\bigl(f(t^kx) - f(t^{k+1}x) - bt^kx\bigr)\right| \leqslant \sum_{k=0}^{n-1}\bigl|f(t^kx) - f(t^{k+1}x) - bt^kx\bigr| < \sum_{k=0}^{n-1} \varepsilon t^k|x|.$$ It follows, by summing the geometric and telescoping series, that \(\displaystyle \left| f(x) - f(t^nx) - \frac{b(1-t^n)x}{1-t} \right| < \frac{\varepsilon(1-t^n)|x|}{1-t}.\) Now you can let $n\to\infty$ and use the facts that $t^nx\to0$ and $f$ is continuous at $0$, to deduce that \(\displaystyle \left| f(x) - f(0) - \frac{bx}{1-t}\right| \leqslant \frac{\varepsilon |x|}{1-t}.\) After dividing through by $|x|$, you should be able to see that $f'(0)$ exists, and is equal to $\dfrac b{1-t}.$

 

[evinda]

ZaidAlyafey's comment shows that if $f'(0)$ exists then it is equal to $\dfrac b{1-t}.$ The snag with this, and with all the other comments, is that you are not told that $f$ is differentiable. You are not even told that it is continuous, except at the single point $0$. So you cannot use l'Hôpital's rule, or the mean value theorem, or any other result that requires differentiability, and you need to prove that $f$ is differentiable at $0$.

You are told that $$\lim_{x\rightarrow 0}\frac{f(x)-f(tx)}{x}=b$$. So given $\varepsilon>0$ there exists $\delta>0$ such that \(\displaystyle \left|\frac{f(x)-f(tx)}{x} - b\right| < \varepsilon\) whenever $|x|<\delta.$ Therefore $$|f(x) - f(tx) - bx| < \varepsilon|x|.$$ Next, if $|x|<\delta$ then $|t^nx|<\delta$ for all $n\geqslant1$ (because $t<1$). So in the above inequality we can replace $x$ by $tx$, $t^2x,\ldots, t^{n-1}x,$ to get $$|f(tx) - f(t^2x) - btx| < \varepsilon t|x|,$$ $$|f(t^2x) - f(t^3x) - bt^2x| < \varepsilon t^2|x|,$$ $$\ldots$$ $$|f(t^{n-1}x) - f(t^nx) - bt^{n-1}x| < \varepsilon t^{n-1}|x|.$$ Now add those inequalities and use the triangle inequality, to get $$\left|\sum_{k=0}^{n-1}\bigl(f(t^kx) - f(t^{k+1}x) - bt^kx\bigr)\right| \leqslant \sum_{k=0}^{n-1}\bigl|f(t^kx) - f(t^{k+1}x) - bt^kx\bigr| < \sum_{k=0}^{n-1} \varepsilon t^k|x|.$$ It follows, by summing the geometric and telescoping series, that \(\displaystyle \left| f(x) - f(t^nx) - \frac{b(1-t^n)x}{1-t} \right| < \frac{\varepsilon(1-t^n)|x|}{1-t}.\) Now you can let $n\to\infty$ and use the facts that $t^nx\to0$ and $f$ is continuous at $0$, to deduce that \(\displaystyle \left| f(x) - f(0) - \frac{bx}{1-t}\right| \leqslant \frac{\varepsilon |x|}{1-t}.\) After dividing through by $|x|$, you should be able to see that $f'(0)$ exists, and is equal to $\dfrac b{1-t}.$

I understand!Thank you very much! :D