calculus - Usage of mean value theorem ; bounded derivative and open interval

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.

Answer

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.