Limits with Taylor series around zero

Question

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 $0$.

Thank you in advance for your help!

Answer 1

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.

Answer 2

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.

Answer 3

Hint:

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