What real $p$ makes this limit exists and be finite?

Question

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.

Answer 1

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?

Answer 2

Render

$\dfrac{\sqrt{1+x}-(1+px)}{x^2}=\dfrac{(\sqrt{1+x}-(1+px))(\sqrt{1+x}+(1+px))}{x^2(\sqrt{1+x}+(1+px))}=\dfrac{(1+x-(1+px)^2)}{x^2(\sqrt{1+x}+(1+px))}$

where you have used the difference of squares factorization $(a+b)(a-b)=a^2-b^2$ in the numerator. That's half the battle, one might say.

Now: The denominator is still dropping off in proportion to $x^2$ and the numerator has to do the same, which is possible only if you put in the right value of $p$ to cancel out the lower degree terms.

Answer 3

Apart from the algebraic methods above, you can use the expansion $$ \sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\dots $$ and then it is clear that when $p\neq 1/2$, a linear term remains in the numerator and your limit is infinity. If $p=1/2$, your limit is just $-1/8$. Taylor's expansions are a much more powerful tool to decide these things, to find limits, etc. Algebraic tricks are fragile and seldom available.