# Limits with Taylor series around zero

I had some problems with the following two limits, which are supposed to be calculated with Taylor series:

$$\lim_{x\to 0^+}\frac{e^\sqrt{x}-e^{-\sqrt{x}}}{\sqrt{\sin{2x}}}\quad\mbox{and}\quad \lim_{x\to 0^+}\frac{(1-\log{x})^{\sin{x^2}}}{(\arctan{x})^{3/2}}.$$

Although the numerators are quite simple to develop in series, I stopped when I noticed that both denominators are not derivable in $$x=0$$, that is, we should not use Taylor series in $$x=0$$ to evaluate this functions around $$0$$. I wonder if is possible to consider right derivatives only, and study the behaviour of the denominators in a right neighborhood of $$0$$.

• A good question (+1). – hamam_Abdallah Jan 21 '19 at 19:12

Hint. You can avoid the problem that $$\sqrt{x}$$ is not differentiable at $$0$$ by considering just the Taylor series of $$e^t=1+t+o(t)$$ and $$\sin(t)=t+o(t)$$ as $$t\to 0$$. Then, as $$x\to 0^+$$, $$\frac{e^\sqrt{x}-e^{-\sqrt{x}}}{\sqrt{\sin{2x}}}=\frac{(1+ \sqrt{x}+o(\sqrt{x}))-(1- \sqrt{x}+o(\sqrt{x}))}{\sqrt{2x+o(x)}}=\frac{\sqrt{x}(2+o(1))}{\sqrt{x}\sqrt{2+o(1)}}.$$ Can you take it from here? Use a similar approach also for the second one.

• Don't you want big O notation? – bounceback Jan 21 '19 at 19:08
• Little-o notation suffices. – Robert Z Jan 21 '19 at 19:09
• Oh, I see what you've done now, yes – bounceback Jan 21 '19 at 19:13
• Thank you @RobertZ, that is what I needed. For the second limit, I have just figured out that the numerator is $1+o(1)$, so it is not necessary to develop the denominator. – Fabio Ori Jan 21 '19 at 20:10

hint for the first

We have $$e^X=1+X(1+\epsilon(X))$$

then $$e^{\sqrt{x}}=1+\sqrt{x}(1+\epsilon(x))$$ and when $$x\to 0^+$$, $$\sqrt{\sin(2x)}\sim \sqrt{2x}$$ thus your limit is

$$\lim_{x\to 0^+}\frac{2\sqrt{x}(1+\epsilon(x))}{\sqrt{2x}}=\sqrt{2}.$$

TOMORROW I WILL LOSE POINTS . NO ONE KNOWS THE REASON.

For this, i leave.

Hint:

$$\dfrac{e^{\sqrt x}-e^{-\sqrt x}}{\sqrt{\sin2x}}=\dfrac{\dfrac{e^{\sqrt x}-1}{\sqrt x}+\dfrac{e^{-\sqrt x}-1}{-\sqrt x}}{\sqrt2\cdot\sqrt{\dfrac{\sin2x}{2x}}}$$

• Thank you! This is an elegant solution, but, if I am not mistaking, it does not make usage of Taylor series. – Fabio Ori Jan 21 '19 at 19:29