What real $p$ makes this limit exists and be finite? The limit is $\lim_{x \to 0} \frac{\sqrt{1+x}-(1+px)}{x^2}$ We must find what real $p$ makes this limit exist and be finite, and determine this limit. I know it can be done using derivative ideas, but the exercise asks not to use any derivative artifice, what I am not being able to develop well.

Make the change: $\sqrt{1+x}=t$. Then: $$\begin{align}\lim_{x \to 0} \frac{\sqrt{1+x}-(1+px)}{x^2}&=\lim_{t \to 1} \frac{t-(1+p(t^2-1))}{(t^2-1)^2}=\\\ &=\lim_{t \to 1} \frac{(t-1)(1-p-pt)}{(t-1)^2(t+1)^2}=\\\ &=\lim_{t \to 1} \frac{1-p-pt}{4(t-1)}.\end{align}$$ For the limit to exist, what must the numerator be equal to? Can you figure out the value of $p$ now?