Find for what values of a and b in R the limit exists (No De L'Hopital) I was given this exercise in my math course at university.
The question is to find, without using De l'Hopitals and other methods which may use derivates and similars, for what values of $a$ and $b$ in $\mathbb{R}$ the following statement is true.
$$
f(x)= \left\{\begin{matrix}  \dfrac{\sin x}{x} & x > 0  \\  ax + b & x \leq 0  \\ \end{matrix}\right. $$ $$ \lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
I started by writing it as:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x} -ax - b}{x} = 0$$ because, due to the fact that we are approaching from $x > 0, \, f(x) = \dfrac{\sin x}{x}$
Then I multiplied $\frac{\sin x}{x}$ by $\frac{\sin x}{\sin x}$ in order to get:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin^2x}{x\sin x} -ax - b}{x} = 0$$
From there:
$$ \lim_{x \to 0^+} \dfrac{\dfrac{\sin^2x}{x\sin x} -ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos^2x}{x\sin x}-ax-b}{x} = 0 $$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x} \times \dfrac{1+\cos x}{\sin x}-ax-b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x} \times \dfrac{1+\cos x}{x} \times \dfrac{x}{\sin x} - ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1 - \cos x}{x} \times \dfrac{1}{x} \times \left(1 + \cos x\right) \times \dfrac{x}{\sin x} -ax - b}{x} = 0$$
$$ \Longleftrightarrow \lim_{x \to 0^+} \dfrac{\dfrac{1-\cos x}{x^2} \times \left(1+ \cos x\right) \times \dfrac{x}{\sin x} -ax - b}{x} = 0$$
Which, substituting the value of $x$ to the value it is approaching, results in:
$$ \dfrac{\dfrac{1}{2}\times\left(1+1\right)\times 1 - a \times 0 - b}{x} = 0 $$
$$ \Longleftrightarrow \dfrac{1 - b}{x} = 0 $$
Which should resolve to the I.F. $$\dfrac{0}{0}$$ when $b=1$
So to me it seems like, with the calculations I've done, I can't reach an answer about which value of $a$ and $b$ make the statement true

Assuming the limit exists:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x} -ax - b}{x} = \lim_{x \to 0^+} \dfrac{\sin x -ax^2 - bx}{x^2}=\lim_{x \to 0^+} \left(\dfrac{\sin x- bx}{x^2}-a\right)
\end{equation}
Now consider:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\sin x- bx}{x^2}
\end{equation}
If $b\neq1$:
\begin{equation}
\lim_{x \to 0^+} \dfrac{\sin x- bx}{x^2}=\lim_{x \to 0^+} \dfrac{\dfrac{\sin x}{x}- b}{x}= \pm \infty
\end{equation}
If $b=1$, we have:
$$ \forall x \in (0, \pi/2), \sin x \leq x \leq \tan x$$
So:
\begin{equation}
\dfrac{\sin x - \tan x}{x^2} \leq \dfrac{\sin x - x}{x^2} \leq 0. 
\end{equation}
Now, considering that:
$$ \sin x - \tan x = \tan x(\cos x - 1) = - \dfrac{\tan x \sin^2 x}{1+\cos x} $$
and
$$\dfrac{\sin x}{x} \to 1 \quad \text{as} \quad x\to 0$$
we have
$$ \lim_{x \to 0} \frac{\sin x - \tan x}{x^2} = - \lim_{x \to 0} \frac{\tan x}{1+\cos x}\left(\frac{\sin x}{x}\right)^2 = 0. $$
Applying the squeezing theorem, we conclude:
$$ \lim_{x\to0^+} \frac{\sin x - x}{x^2} = 0. $$
Since
$$ \lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
and
$$ \lim_{x \to 0^+} \left(\dfrac{\sin x- bx}{x^2}-a\right)=\lim_{x \to 0^+} \dfrac{f(x) - ax - b}{x} = 0$$
we must have $a=0$.
