Calculating $\displaystyle{\lim_{x \to 0^+}}{\frac{1}{\sqrt{x}}\Big(e^x + \frac{2\log(\cos(x))}{x^2}}\Big)$ I am struggling to calculate this limit:
$$\displaystyle{\lim_{x \to 0^+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}}$$
I prefer not to use l'Hopital's rule, only when necessary. If possible, solving with the help of these limits:
$\displaystyle{\lim_{x \to 0^+}}{\frac{\log(x + 1)}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{e^x - 1}{x}} = 1$, $\displaystyle{\lim_{x \to 0^+}}{\frac{1 - \cos(x)}{x^2}} = \frac{1}{2}$.
Also when I tried solving it using only l'Hopital, it seemed to be very laborious, and also not sure if sufficient to solve it.
Also I prefer not using Taylor theorem and little/big o notation if possible at all.
Here's my (unsuccessful) attempt:
$$\displaystyle{\lim_{x \to 0+}}{\frac{e^x + \frac{2\log(\cos(x))}{x^2}}{\sqrt{x}}} = \displaystyle{\lim_{x \to 0^+}}{\frac{e^xx^2 + e^x2x + 2\log{\cos{x}}}{x^2\sqrt{x}}} = \frac{1}{5}\displaystyle{\lim_{x \to 0^+}}{\frac{e^x(x^2 + 2x) - 2tg{x}}{x\sqrt{x}}} = ...$$
Problem is, each time I use l'Hopital, it doesn't seem to simplify limit in any way.
Thanks.

As the OP asks, an attempt to do it, using only
\begin{align}
\log(1+u)&=u+o(u),\\
e^x &= 1 + x + o(x),\\
\cos(x) &= 1 -\frac{x^2}{2} + o(x^2) .
\end{align}
Now, as $x \to 0$, the best we can deduce is:
\begin{align}
\cos x &= 1 - \frac{x^2}{2} + o(x^2)
\\
\log \cos x &= \log(1-(1-\cos x)) = -(1-\cos x) + o\big(1-\cos x\big)
\\ &= -1+\cos x + o\big(x^2\big)
= -1 + \big(1-\frac{x^2}{2} + o(x^2)\big)+ o\big(x^2\big)
\\ &=-\frac{x^2}{2}+o(x^2)
\\
\frac{2\log \cos x }{x^2} &= -1+o(1)
\\
e^x &= 1+x+o(x)
\\
e^x+\frac{2\log \cos x }{x^2} &= 1+x+o(x)-1+o(1) =o(1)
\\
\frac{e^x+\frac{2\log \cos x }{x^2}}{\sqrt{x}} &= o(x^{-1/2})
\end{align}
This is not enough to get the answer.
From this we can see how to get a counterexample.  Instead of $\cos(x)$, use $c(x) := 1-\frac{x^2}{2}+x^{5/2}$.  This still satisfies $c(x) = 1-\frac{x^2}{2}+o(x^2)$, but
$$
\frac{e^x+\frac{2\log c(x) }{x^2}}{\sqrt{x}} \to 2
$$
and not $0$.
