Value of the limit at $x=0$ What is $$\lim_{x \to  0^+} \left((x\cos(x))^x+(x\sin(x)\right)^{1/x}?$$ I substituted $x = 0+h$ and used that near $0, \cos(h) \approx 1, \sin(h)\approx h$ to get the new limit as $$\lim_{h \to 0} \left(h^h+h^{2/h}\right).$$
But now the problem is the second term. How to reduce it to form like $0/0,\infty/\infty$? Thanks.
 A: Hint. As $x \to 0^+$, one may use the Taylor series expansion of $\sin x$, getting
$$
\begin{align}
\sin (x)&=x+O(x^3),
\\\\ \log(\sin (x))&=\log\left(x+O(x^3)\right)=\log x+O(x^2),
\\\\ \frac{\log(\sin (x))}x&=\frac{\log x}x+O(x),
\end{align}
$$ giving, as $x \to 0^+$,
$$
\begin{align}
\left(x\cos(x))^x+(x\sin(x)\right)^{1/x}&=e^{\large x\log x + x\log\left(\cos x \right)}+e^{\large \frac{\log x}x + \large \frac{\log\left(\sin x \right)}x}
\\\\ &\to e^{0+0}+0
\\\\ &\to 1.
\end{align}
$$
A: There is a mismatch of parentheses in the question as asked. One possible correction is
$$
\begin{align}
\lim_{x\to0}\left((x\cos(x))^x+x\sin(x)\right)^{1/x}
&=\lim_{x\to0}x\cos(x)\cdot\lim_{x\to0}\left(1+\frac{x\sin(x)}{(x\cos(x))^x}\right)^{1/x}\\
&=0\cdot1
\end{align}
$$
Note that $\lim\limits_{x\to0}\left(1+\frac{x\sin(x)}{(x\cos(x))^x}\right)^{1/x}$ is between $\lim\limits_{x\to0}\left(1+ax\right)^{1/x}$ and $\lim\limits_{x\to0}\left(1-ax\right)^{1/x}$ for any positive $a$.

Another is
$$
\begin{align}
\lim_{x\to0}\left((x\cos(x))^x+(x\sin(x))^{1/x}\right)
&=\lim_{x\to0}x^x\cdot\lim_{x\to0}\cos(x)^x+\lim_{x\to0}x^{1/x}\cdot\lim_{x\to0}\sin(x)^{1/x}\\
&=1\cdot1+0\cdot0
\end{align}
$$
For $\lim\limits_{x\to0}x^x$, see this answer. The other limits are easier.
A: First, a good idea is to use WolframAlpha to get an idea of what the solution should be (and possibly also an idea of how to solve it).
Here's the result of WolframAlpha as image: WolframAlpha Query
So, the result is $1$. But how do we get to it? We will use L'Hospital in a different way than you thought. First, let's simplify a bit:
$$\lim_{x \to 0^+} \left((x \cos(x))^x + (x \sin(x))^{\frac{1}{x}}\right) = \lim_{x \to 0^+} (x \cos(x))^x + \lim_{x \to 0^+} (x \sin(x))^{\frac{1}{x}}$$
Lets now compute the first expression. We want to somehow extract the exponent from the expression. This can be done with the following trick:
$$a^b = e^{\log(a^b)} = e^{b \cdot \log(a)}$$
The trick works for real numbers as $\log$ is the inverse function of $e$, not for complex (there $\log$ is not bijective).
$$(x \cos(x))^x = e^{\log((x \cos(x))^x)} = e^{x \cdot \log(x \cos(x))}$$
Okay, now we want to know what $\lim\limits_{x \to 0} x \cdot \log(x \cos(x))$ is. Let's do a little trick and then use L'Hospital.
$$\lim_{x \to 0} x \cdot \log(x \cos(x)) = \lim_{x \to 0} \frac{\log(x \cos(x))}{\frac{1}{x}} \stackrel{LH}{=} \lim_{x \to 0} \frac{\frac{1}{x \cos(x)} \cdot \left(1 \cdot \cos(x) + x \cdot (-\sin(x))\right)}{-\frac{1}{x^2}}\\
= \lim_{x \to 0} -x^2 \frac{\cos(x) - x \sin(x)}{x \cos(x)} = \lim_{x \to 0} -x^2 \frac{1}{x} \left(\frac{\cos(x)}{\cos(x)} - \frac{x \sin(x)}{\cos(x)}\right)\\
= \lim_{x \to 0} -x \left(1 - x \tan(x)\right) = 0$$
Well, if we go back we now have $e^0 = 1$, thus the left expression converges to $1$.
$$\lim_{x \to 0^+} (x \cos(x))^x = 1$$
Last we need to show that the right expression converges to $0$. We will use a similar approach.
$$(x \sin(x))^{\frac{1}{x}} = e^{\frac{1}{x} \log(x \sin(x))}$$
Again, we are interested what the value for $\exp$ does when converging to $0$. But this time the direction is important ($-\infty$ from right and $\infty$ from left). Also remember, by the product rule, the limit of a product is the product of the limits if neither approaches to $0$.
$$\lim_{x \to 0^+} \frac{\log(x \sin(x))}{x} = \lim_{x \to 0^+} \frac{1}{x} \cdot \lim_{x \to 0^+} \log(x \sin(x)) = -\infty$$
The later is true since informally $\log$ (as $\exp$) converges faster than other functions. You may google for a formal description and proof if you're not familiar with that.
Okay, we now have $e^{-\infty}$ for the second part, which converges to $0$.
In total we have:
$$\lim_{x \to 0^+} \left((x \cos(x))^x + (x \sin(x))^{\frac{1}{x}}\right) = \lim_{x \to 0^+} (x \cos(x))^x + \lim_{x \to 0^+} (x \sin(x))^{\frac{1}{x}} = 1 + 0 = 1$$
And are finished. I hope you understood everything :)
