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Let $f : (0,1) \to \mathbb{R}$ be a function such that $ |f'(x)| \leq 5 $ on the open interval $(0,1)$. Prove that $\lim_{x \to 1^-} f(x)$ exists.

It involves the derivative and the actual function itself, so I think I have to somehow use the mean value theorem.. Also, $f$ is continuous on $(0,1)$ and differentiable on $(0,1)$ ( because the derivative exists there ).

But then, the function is defined on the open interval, so the requirements for the mean value theorem aren't satisfied. I'm guessing we have to consider intervals of the form $(a,b)$ with $a > 0$ and $b < 0$.

Lastly, I don't see the significance of the $5$ ... Is it only there to establish that the derivative is bounded, or does the number itself have some signifiance ( would the same thing hold if we had $3$ for example? ).

Please give me a hint, not the solution. Something like "consider the mean value theorem on intervals of the form ... " would be very helpful.

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    $\begingroup$ Right, the $5$ isn't important - just the fact that the derivative is bounded. The point is that if you draw the graph of $f$ and choose a point $(x_0, f(x_0))$ on it, then the graph is stuck inside a cone whose vertex is at that point. You can prove this with the mean value theorem, and once you draw the picture of this, you should have the intuition needed to complete the proof. $\endgroup$ – user296602 Jan 6 '18 at 22:26
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Pick a sequence $(x_{n})\subseteq(0,1)$ such that $x_{n}\rightarrow 1$. Then \begin{align*} |f(x_{n})-f(x_{m})|=|f'(\eta_{n,m})||x_{n}-x_{m}|\leq 5|x_{n}-x_{m}|, \end{align*} where $\eta_{n,m}$ is chosen by Mean Value Theorem. So $(f(x_{n}))$ is convergent. For other sequence $(y_{n})$ such that $y_{n}\rightarrow 1$, consider the sequence $(z_{n})$ defined by $z_{2n}=x_{n}$, $z_{2n+1}=y_{n}$ to claim that the limits of $(f(x_{n}))$ and $(f(y_{n}))$ are the same. So $\lim_{x\rightarrow 1^{-}}f(x)$ exists.

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Hint: using MVT shows that for every sequence converging to $1$, $(f(x_n))$ is a Cauchy sequence (since $|f(x_n)-f(x_m)|\leq 5|x_n-x_m|$) and thus has a limit, MVT again will enables you to show that if $lim_ny_n=1$, $lim_nf(x_n)=lim_nf(y_n)$. $(|f(x_n)-f(y_n)|\leq 5|x_n-y_n|)$ so $lim_nf(x_n)=lim_nf(y_n)$.

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Let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $(0,1)$ such that $\lim_{n\in\mathbb N}x_n=1$. The the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ is a Cauchy sequence. Indeed, given $\varepsilon>0$, there is a $p\in\mathbb N$ suh that$$m,n\geqslant p\implies|x_m-x_n|<\frac\varepsilon5$$and therefore$$m,n\geqslant p\implies\bigl|f(x_m)-f(x_n)\bigr|\leqslant5|x_m-x_n|<\frac\varepsilon5;$$this is where $|f'|<5$ is used.

So $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ converges to som $l$. In fact, all such sequence converge converges to the same number, independently of the choice of the sequence. Indeed, if $(y_n)_{n\in\mathbb N}$ is another sequence of elements of $(0,1)$ such that $\lim_{n\in\mathbb N}y_n=1$ and if $\lim_{n\to\infty}f(y_n)=l'\neq l$, then the seqeunce$$x_1,y_1,x_2,y_2,x_3,y_3,\ldots$$would converge to $1$, but the sequence$$f(x_1),f(y_1),f(x_2),f(y_2),f(x_3),f(y_3),\ldots$$ would not converge and I have already proved that this cannot happen.

It is now easy to prove that $\lim_{x\to1^-}f(x)=l.$

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Let $a, x$ be such that $0<a<x<1$ then by mean value theorem $$f(x) =f(a) +(x-a) f'(c) $$ Since $f'$ is bounded it follows that $\limsup_{x\to 1^{-}}f(x)=A, \liminf_{x\to 1^{-}} f(x) =B$ exist as real numbers. If $A\neq B$ then we can find values $x, x'$ near to and less than $1$ such that $f(x) $ is near $A$ and $f(x') $ is near $B$ and further $|f(x) - f(x') |>(A-B) /2$ and thus the ratio $(f(x) - f(x')) /(x-x')=f'(c) $ becomes unbounded as $x, x'$ get sufficiently close to $1$. This contradiction shows that $A=B$ and hence $\lim_{x\to 1^{-}}f(x)$ exists.

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