$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$ 
Find  $$\lim_{x \to 0}\frac{|x|\sin \left(\frac{1}{3 \sqrt{x}}\right)}{\sqrt{x^4+4x^2+7}}$$

I know that $\lim_{x \to 0} \frac{\sin x}{x}=1$ But here $\sin \left(\frac{1}{3 \sqrt{x}}\right)$ is given when $x \to 0$. Need help.
 A: The limit is zero, because $|x| \sin{\frac{1}{3 \sqrt{x}}} \rightarrow 0$ as $x \rightarrow 0$.  (The denominator is nonzero in this limit.). 
A: The limit of the denominator is $\sqrt7$ so we just need to tame the numerator . Observe
$$\left|\left|x\right|\sin\frac{1}{3\sqrt{x}}\right|=\left|x\right|\left|\sin\frac{1}{3\sqrt{x}}\right|\le \left|x\right|\cdot 1$$
What does this tell you?
Also note that $x\to 0^+$ as $\sqrt x$ must be defined
A: You may want to prove the easy and pretty useful 
Lemma: If $\,f(x)\xrightarrow [x\to x_0]{}0\,$ and $\,|g(x)|\leq M\,\,\,\forall\,x\in(x_0-\epsilon\,,\,x_0+\epsilon)\,$ , for some $\,\epsilon>0\,$ , then
$$\lim_{x\to x_0}f(x)g(x)=0$$
The above simply says that the limit of a function converging to zero times a bounded function is zero.
Now, since $\,\displaystyle{\left|\sin\frac{1}{3\sqrt x}\right|\leq 1\,\,,\,\,x>0}\,$, we get at once, applying the above lemma to the numerator:
$$\frac{x\sin\frac{1}{3\sqrt x}}{\sqrt{x^4+4x^2+7}}\xrightarrow [x\to 0^+]{}\frac{0}{\sqrt 7}=0$$
Note that as Tetori wrote, your function's defined only for positive $\,x\,$, rendering the absolute value in the numerator useless.
A: If we let $1/x=y$
$$\lim_{y \to \infty}\frac{\sin \left(\frac{\sqrt{y}}{3 }\right)}{|y|\sqrt{7}}=0$$
